Quadratic equations have the general form $latex ax^2+bx+c$. These equations can have two real solutions, one real solution, or no real solution. There are several methods we can use to solve quadratic equations. Some of the most important methods are the factoring method, the method of completing the square, and the general quadratic formula.

Here, we will look at a summary of the methods we can use to solve quadratic equations. Then, we will look at 10 quadratic equations examples with solutions.

##### ALGEBRA

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Learning to solve quadratic equations with examples.

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##### ALGEBRA

Relevant for

Learning to solve quadratic equations with examples.

See examples

## Summary of how to solve quadratic equations

Quadratic equations are characterized by their variables having a maximum power of 2. These equations have the general form $latex ax^2+bx+c=0$. For example, the equations $latex 5x^2+2x+4=0$ and $latex 4x^2-5x-5=0$ are quadratic equations.

We can solve these types of equations using different methods, depending on the quadratic equation we have. For example, we have methods for solving incomplete quadratic equations, the factoring method, the completing the square method, and the general quadratic formula.

To solve incomplete quadratic equations of the form $latex ax^2+c=0$, we start by isolating x² completely. Then, we need to take the square root of both sides of the equation.

To solve incomplete quadratic equations of the form $latex ax^2+bx=0$, we start by factoring the x from both terms. Then, we have to get a linear equation with each factor and solve for x.

To solve complete quadratic equations of the form $latex ax^2+bx+c=0$, we can use the factorization method. For this, we have to write the equation in the form $latex (x+p)(x+q)=0$ and form a linear equation with each factor.

Finally, we have the general quadratic formula that allows us to solve any quadratic equation and obtain its two roots:

$$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$

If you need to learn or review the methods of solving quadratic equations, you can visit our article: Solving Quadratic Equations – Methods and Examples.

The following 10 quadratic equation examples are solved using various quadratic equation solving methods. Try to solve the problems yourself before looking at the answer.

### EXAMPLE 1

Solve the equation $latex x^2-16=0$.

This quadratic equation does not have a bx term, that is, it has the form $latex ax^2+c=0$. To solve this equation, we can isolate the x² term and take the square root of both sides of the equation:

$latex x^2-25=0$

$latex x^2=25$

Now, we take the square root of both sides of the equation and we have:

$latex x=\pm \sqrt{25}$

$latex x=\pm 5$

The solutions of the equation are $latex x=5$ and $latex x=-5$.

### EXAMPLE 2

Find the solutions of the equation $latex x^2-9x=0$.

This quadratic equation does not have a c term, that is, it has the form $latex ax^2+bx=0$. We can solve this equation by factoring x and then forming an equation with each factor:

$latex x^2-9x=0$

$latex x(x-9)=0$

Now, we can obtain the following linear equations and solve:

$latex x=0~~$ or $latex ~~x-9=0$

$latex x=0~~$ or $latex ~~x=9$

The solutions of the equation are $latex x=0$ and $latex x=9$.

### EXAMPLE 3

Find the roots of the equation $latex x^2-6x-7=0$.

Let’s solve this equation using factorization. Therefore, we have to look for two numbers, which when multiplied result in -7, and when added result in -6.

The numbers we are looking for are -7 and 1. Then, we have:

$latex x^2-6x-7=0$

$latex (x-7)(x+1)=0$

Now, we can form a linear equation with each factor:

$latex x-7=0~~$ or $latex ~~x+1=0$

$latex x=7~~$ or $latex ~~x=-1$

The solutions of the equation are $latex x=7$ and $latex x=-1$.

### EXAMPLE 4

Solve the equation $latex x^2-8x+4=0$. Express the solutions to two decimal places.

To solve this equation, we can use the quadratic formula. Therefore, we use the coefficients $latex a=1$, $latex b=-8$ and $latex c=4$ and we have:

$$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$

$$x=\frac{-(-8)\pm \sqrt{( -8)^2-4(1)(4)}}{2(1)}$$

$$=\frac{8\pm \sqrt{64-16}}{2}$$

$$=\frac{8\pm \sqrt{48}}{2}$$

$$x=7.46 \text{ or } 0.54$$

The solutions are $latex x=7.46$ and $latex x=0.54$.

### EXAMPLE 5

Solve the equation $latex -x^2-5x+25=-2x^2+5x$ using the quadratic formula.

To use the quadratic formula, we have to start by writing the equation in the form $latex ax^2+bx+c=0$. Therefore, we have:

$latex -x^2-5x+25=-2x^2+5x$

$latex x^2-10x+25=0$

Now, we have the coefficients $latex a=1$, $latex b=-10$, and $latex c=25$. Using them in the general quadratic formula, we have:

$$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$

$$x=\frac{-(-10)\pm \sqrt{( -10)^2-4(1)(25)}}{2(1)}$$

$$=\frac{10\pm \sqrt{100-100}}{2}$$

$$=\frac{10\pm \sqrt{0}}{2}$$

$$=\frac{10}{2}$$

$latex x=5$

In this case, we have a single repeated root $latex x=5$.

### EXAMPLE 6

Solve the equation $latex 4x^2+5=2x^2+20$.

To solve the equation, we have to start simplifying and write it in the form $latex ax^2+bx+c=0$:

$latex 4x^2+5=2x^2+20$

$latex 4x^2-2x^2+5-20=0$

$latex 2x^2-15=0$

Now, we can solve by isolating x² and taking the square root of both sides:

$latex 2x^2-15=0$

$latex 2x^2=15$

$latex x^2=\frac{15}{2}$

$latex x=\pm \sqrt{\frac{15}{2}}$

### EXAMPLE 7

Find the solutions of the following equation using any method $$(3x+1)(2x-1)-(x+2)^2=5$$

First, we need to expand the parentheses to simplify the equation down to the form $latex ax^2+bx+c=0$. Therefore, we have:

$$(3x+1)(2x-1)-(x+2)^2=5$$

$$6x^2-x-1-(x^2+4x+4)=5$$

$latex 5x^2-5x-5=5$

$latex 5x^2-5x-10=0$

Now, we can solve by factorization:

$latex 5x^2-5x-10=0$

$latex 5(x^2-x-2)=0$

$latex 5(x+1)(x-2)=0$

$latex x+1=0~~$ or $latex ~~x-2=0$

$latex x=-1~~$ or $latex ~~x=2$

### EXAMPLE 8

Use the general quadratic formula or any other method to show that the equation $latex 5x^2+4x+10=0$ has no real solutions.

We are going to use the quadratic formula with the coefficients $latex a=5$, $latex b=4$ and $latex c=10$:

$$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$

$$x=\frac{-(4)\pm \sqrt{( 4)^2-4(5)(10)}}{2(5)}$$

$$=\frac{-5\pm \sqrt{16-200}}{10}$$

$$=\frac{-5\pm \sqrt{-184}}{10}$$

By simplifying the expression inside the square root, we can see that we got a negative value. $latex \sqrt{-184}$ is not a real number, so the equation has no real roots.

However, if we use imaginary numbers, the equation does have two complex roots.

### EXAMPLE 9

Find the solutions of the following equation $$\frac{2x+1}{x+5}=\frac{3x-1}{x+7}$$

We can start by cross multiplying to simplify the fractions:

$$(2x+1)(x+7)=(3x-1)(x+5)$$

If we expand the parentheses and simplify, we can form a quadratic equation of the form $latex ax^2+bx+c=0$:

$$2x^2+15x+7=3x^2+14x-5$$

$latex x^2-x-12=0$

Solving by factorization, we have:

$latex (x+3)(x-4)=0$

$latex x+3=0~~$ or $latex ~~x-4=0$

$latex x=-3~~$ or $latex ~~x=4$

### EXAMPLE 10

We have two numbers that have a sum equal to 17 and when we multiply them, we get 60. Find the two numbers.

We can solve this problem by forming equations using the given information. If we use the letters X (smaller number) and Y (larger number) to represent the numbers, we have:

$latex X+Y=17~~[1]$

$latex XY=60~~[2]$

Taking equation 1 and writing as $latex Y=17-X$, we can substitute it into the second equation:

$latex X(17-X)=60$

Now, we can expand and write the equation in the form $latex ax^2+bx+c=0$:

$latex 17X-X^2=60$

$latex X^2-17X+60=0$

Solving by factorization, we have:

$latex X^2-17X+60=0$

$latex (X-12)(X-5)=0$

$latex X-12=0~~$ or $latex ~~X-5=0$

$latex X=12~~$ or $latex ~~X=5$

Therefore, we have:

• If $latex X=5$, we have $latex Y=17-5=12$. This solution is correct because X<Y.
• If $latex X=12$, we have $latex Y=17-12=5$

The numbers are 12 and 5.

## 5 Quadratic equations practice problems

Solve the following problems using any method of solving quadratic equations.