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

Aug 13, 2016

$x = \left\{0 , - \frac{5}{2}\right\}$

#### Explanation:

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

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

$a = 2 \text{ ; "b=5" ; } c = 0$

$\Delta = \sqrt{{b}^{2} - 4 \cdot a \cdot c}$

$\Delta = \sqrt{{5}^{2} - 4 \cdot 2 \cdot 0}$

$\Delta = {\sqrt{5}}^{2}$

Delta=±5

x_("1,2")=(-b±Delta)/(2*a)

${x}_{1} = \frac{- 5 - 5}{2 \cdot 2} = - \frac{10}{4} = - \frac{5}{2}$

${x}_{2} = \frac{- 5 + 5}{2 \cdot 2} = 0$

$x = \left\{0 , - \frac{5}{2}\right\}$

$\text{Or simply}$

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

$2 \cancel{{x}^{2}} = - 5 \cancel{x}$

$2 x = - 5$

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

Aug 13, 2016

$x = - \frac{5}{2} , 0$

#### Explanation:

$2 {x}^{2} + 5 x = 0$ is a quadratic equation in the form $a {x}^{2} + b x + c$, where $a = 2$, $5 = b$, and $c = 0$.

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

Plug the given values into the formula.

$x = \frac{- 5 \pm \sqrt{{\left(- 5\right)}^{2} - 4 \cdot 2 \cdot 0}}{2 \cdot 2}$

Simplify.

$x = \frac{- 5 \pm \sqrt{25 - 0}}{4}$

Simplify.

$x = \frac{- 5 \pm \sqrt{25}}{4}$

Simplify.

$x = \frac{- 5 \pm 5}{4}$

Solve for ${x}_{1}$ and ${x}_{2}$.

$\textcolor{b l u e}{{x}_{1} = \frac{- 5 + 5}{4}}$

$\textcolor{b l u e}{{x}_{1} = \frac{0}{4}}$

$\textcolor{b l u e}{{x}_{1} = 0}$

$\textcolor{red}{{x}_{2} = \frac{- 5 - 5}{4}}$

$\textcolor{red}{{x}_{2} = - \frac{10}{4}}$

$\textcolor{red}{{x}_{2} = - \frac{5}{2}}$

$\textcolor{p u r p \le}{x = - \frac{5}{2} , 0}$