# How do you solve x^2+7x=-3 using the quadratic formula?

Jul 5, 2015

#### Answer:

Rewrite the given formula in standard form, then apply the quadratic formula for roots to get $x = - \frac{1}{2}$ or $x = - \frac{13}{2}$

#### Explanation:

${x}^{2} + 7 x = - 3$
$\textcolor{w h i t e}{\text{XXXX}}$is equivalent to
${x}^{2} + 7 x + 3 = 0$
$\textcolor{w h i t e}{\text{XXXX}}$which is a quadratic in standard form
$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{w h i t e}{\text{XXXX}}$$a {x}^{2} + b x + c = 0$
$\textcolor{w h i t e}{\text{XXXX}}$whose roots are given by the quadratic formula:
$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{w h i t e}{\text{XXXX}}$$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

Using $a = 1$, $b = 7$, and $c = 3$
we have
$\textcolor{w h i t e}{\text{XXXX}}$x= (-7+-sqrt(7^2-4(1)(3)))/(2(1)

$\textcolor{w h i t e}{\text{XXXX}}$$x = \frac{- 7 \pm \sqrt{36}}{2}$

$\textcolor{w h i t e}{\text{XXXX}}$$x = \frac{- 7 \pm 6}{2}$

$x = - \frac{1}{2}$ or $x = - \frac{13}{2}$