# How do you solve 2x^2+5x=0 using the quadratic formula?

Jan 29, 2017

Express in the form $a {x}^{2} + b x + c = 0$ then apply the formula to find:

$x = 0 \text{ }$ or $\text{ } x = - \frac{5}{2}$

#### Explanation:

Given:

$2 {x}^{2} + 5 x = 0$

we can express this equation as:

$2 {x}^{2} + 5 x + 0 = 0$

This is in the form:

$a {x}^{2} + b x + c = 0$

with $a = 2$, $b = 5$ and $c = 0$

It has roots given by the quadratic formula:

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

$\textcolor{w h i t e}{x} = \frac{- 5 \pm \sqrt{{\left(- 5\right)}^{2} - 4 \left(2\right) \left(0\right)}}{2 \cdot 2}$

$\textcolor{w h i t e}{x} = \frac{- 5 \pm \sqrt{25}}{4}$

$\textcolor{w h i t e}{x} = \frac{- 5 \pm 5}{4}$

So one root is:

$x = \frac{- 5 + 5}{4} = 0$

and the other is:

$x = \frac{- 5 - 5}{4} = - \frac{10}{4} = - \frac{5}{2}$

$\textcolor{w h i t e}{}$
Footnote

You really would not normally choose to use the quadratic formula in this example.

Instead note that both terms are divisible by $x$, so it factors like this:

$0 = 2 {x}^{2} + 5 x = x \left(2 x + 5\right)$

Which has solutions $x = 0$ and $x = - \frac{5}{2}$