# How do you solve 3x^2+8=12x by using the Quadratic Formula?

Jul 21, 2015

$x = 2 \pm \frac{4}{\sqrt{3}}$
$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{w h i t e}{\text{XXXX}}$(solved using the Quadratic Formula)

#### Explanation:

$\textcolor{w h i t e}{\text{XXXX}}$$a {x}^{2} + b x + c = 0$
the quadratic formula for the solutions is
$\textcolor{w h i t e}{\text{XXXX}}$$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

It will be necessary, first to convert the given equation:
$\textcolor{w h i t e}{\text{XXXX}}$$3 {x}^{2} + 8 = 12 x$
into the standard form (by subtracting $\left(12 x\right)$ from both sides):
$\textcolor{w h i t e}{\text{XXXX}}$$3 {x}^{2} - 12 x + 8 = 0$

Therefore, using:
$\textcolor{w h i t e}{\text{XXXX}}$$a = 3$$\textcolor{w h i t e}{\text{XXXX}}$$b = - 12$$\textcolor{w h i t e}{\text{XXXX}}$$c = 8$

$\textcolor{w h i t e}{\text{XXXX}}$$x = \frac{12 \pm \sqrt{{\left(- 12\right)}^{2} - 4 \left(3\right) \left(8\right)}}{2 \left(3\right)}$
$\textcolor{w h i t e}{\text{XXXX}}$$x = \frac{12 \pm \sqrt{144 - 96}}{6}$
$\textcolor{w h i t e}{\text{XXXX}}$$x = 2 \pm \frac{4}{\sqrt{3}}$