# How do you solve the quadratic equation by completing the square: v^2+6v-59=0?

Jul 23, 2015

$^ {v}^{2} + 6 v - 59 = 0$
$\textcolor{w h i t e}{\text{XXXX}}$$\rightarrow$$\textcolor{w h i t e}{\text{XXXX}}$$v = - 3 \pm 2 \sqrt{17}$
$\textcolor{w h i t e}{\text{XXXX}}$(by completing the square)

#### Explanation:

${v}^{2} + 6 v - 59 = 0$
$\textcolor{w h i t e}{\text{XXXX}}$Move the constant to the right side (out of the way)
${v}^{2} + 6 v = 59$
$\textcolor{w h i t e}{\text{XXXX}}$Add as the third term the value needed to make the right side a square
${v}^{2} + 6 v + {3}^{2} = 59 + 9$
$\textcolor{w h i t e}{\text{XXXX}}$Re-write right side as a squared binomial
${\left(v + 3\right)}^{2} = 68$
$\textcolor{w h i t e}{\text{XXXX}}$Take the square root of both sides
$v + 3 = \pm \sqrt{68} = \pm 2 \sqrt{17}$
$\textcolor{w h i t e}{\text{XXXX}}$Isolate $v$ by subtracting $3$ from both sides
$v = - 3 \pm 2 \sqrt{17}$