# How do you solve x^2+8x+15=0  using the quadratic formula?

Mar 20, 2018

See a solution process below:

#### Explanation:

For $\textcolor{red}{a} {x}^{2} + \textcolor{b l u e}{b} x + \textcolor{g r e e n}{c} = 0$, the values of $x$ which are the solutions to the equation are given by:

$x = \frac{- \textcolor{b l u e}{b} \pm \sqrt{{\textcolor{b l u e}{b}}^{2} - \left(4 \textcolor{red}{a} \textcolor{g r e e n}{c}\right)}}{2 \cdot \textcolor{red}{a}}$

Substituting:

$\textcolor{red}{1}$ for $\textcolor{red}{a}$

$\textcolor{b l u e}{8}$ for $\textcolor{b l u e}{b}$

$\textcolor{g r e e n}{15}$ for $\textcolor{g r e e n}{c}$ gives:

$x = \frac{- \textcolor{b l u e}{8} \pm \sqrt{{\textcolor{b l u e}{8}}^{2} - \left(4 \cdot \textcolor{red}{1} \cdot \textcolor{g r e e n}{15}\right)}}{2 \cdot \textcolor{red}{1}}$

$x = \frac{- \textcolor{b l u e}{8} \pm \sqrt{64 - 60}}{2}$

$x = \frac{- \textcolor{b l u e}{8} - \sqrt{4}}{2}$ and $x = \frac{- \textcolor{b l u e}{8} + \sqrt{4}}{2}$

$x = \frac{- \textcolor{b l u e}{8} - 2}{2}$ and $x = \frac{- \textcolor{b l u e}{8} + 2}{2}$

$x = \frac{- 10}{2}$ and $x = \frac{- 6}{2}$

$x = - 5$ and $x = - 3$

The Solution Set Is: $x = \left\{- 5 , - 3\right\}$

Mar 20, 2018

$x = - 5$ or $x = - 3$

#### Explanation:

Quadratic formula gives the solution of a quadratic equation. For a quaddratic equation $a {x}^{2} + b x + c = 0$, the roots given by quadratic formula are $\frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$.

Hence for ${x}^{2} + 8 x + 15 = 0$, we have $a = 1$, $b - 8$ and $c = 15$ and hence solution is

$\frac{- 8 \pm \sqrt{{8}^{2} - 4 \cdot 1 \cdot 15}}{2 \cdot 1}$

= $\frac{- 8 \pm \sqrt{64 - 60}}{2}$

= $\frac{- 8 \pm \sqrt{4}}{2}$

= $\frac{- 8 \pm 2}{2}$

i.e. either $x = - 5$ or $x = - 3$