# How do you solve 7x^2-5=2x+9x^2 using the quadratic formula?

Aug 10, 2017

$x = - \frac{1 + 3 i}{2} ,$ $x = - \frac{1 - 3 i}{2}$

Refer to the explanation for the process.

#### Explanation:

First combine like terms and rewrite the equation in standard form.

$7 {x}^{2} - 5 = 2 x + 9 {x}^{2}$

Move all terms to the left side.

$7 {x}^{2} - 5 - 2 x - 9 {x}^{2} = 0$

Simplify.

$- 2 {x}^{2} - 5 - 2 x - 9 {x}^{2} = 0$

Rewrite in standard form: $a {x}^{2} + b x + c = 0$.

$- 2 {x}^{2} - 2 x - 5 = 0$,

where $a = - 2$, $b = - 2$, and $c = - 5$.

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

Plug in the known values.

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

Simplify.

$x = \frac{2 \pm \sqrt{- 36}}{- 4}$

Simplify.

$x = - \frac{2 \pm 6 i}{4}$

$x = - \frac{2 + 6 i}{4} ,$ $\frac{- 2 - 36 i}{4}$

Reduce.

$x = - \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{{2}^{1}}}} + \textcolor{red}{\cancel{\textcolor{b l a c k}{{6}^{3}}}} i}{\textcolor{red}{\cancel{\textcolor{b l a c k}{{4}^{2}}}}} ,$ $- \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{{2}^{1}}}} - \textcolor{red}{\cancel{\textcolor{b l a c k}{{6}^{3}}}} i}{\textcolor{red}{\cancel{\textcolor{b l a c k}{{4}^{2}}}}} ,$

$x = - \frac{1 + 3 i}{2} ,$ $x = - \frac{1 - 3 i}{2}$