# How do you solve 2x^2 = 3x +9 by completing the square?

Mar 6, 2017

See the entire solution process below:

#### Explanation:

First, subtract $\textcolor{red}{3 x}$ and $\textcolor{b l u e}{9}$ from each side of the equation to put the quadratic equation in standard form:

$2 {x}^{2} - \textcolor{red}{3 x} - \textcolor{b l u e}{9} = 3 x + 9 - \textcolor{red}{3 x} - \textcolor{b l u e}{9}$

$2 {x}^{2} - 3 x - 9 = 3 x - \textcolor{red}{3 x} + 9 - \textcolor{b l u e}{9}$

$2 {x}^{2} - 3 x - 9 = 0 + 0$

$2 {x}^{2} - 3 x - 9 = 0$

$\left(2 x + 3\right) \left(x - 3\right) = 0$

Now, solve each term for $0$:

Solution 1)

$2 x + 3 = 0$

$2 x + 3 - \textcolor{red}{3} = 0 - \textcolor{red}{3}$

$2 x + 0 = - 3$

$2 x = - 3$

$\frac{2 x}{\textcolor{red}{2}} = - \frac{3}{\textcolor{red}{2}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} x}{\cancel{\textcolor{red}{2}}} = - \frac{3}{2}$

$x = - \frac{3}{2}$

Solution 2)

$x - 3 = 0$

$x - 3 + \textcolor{red}{3} = 0 + \textcolor{red}{3}$

$x - 0 = 3$

$x = 3$

$\frac{2 x}{\textcolor{red}{2}} = - \frac{3}{\textcolor{red}{2}}$

The solution is: $x = - \frac{3}{2}$ and $x = 3$