# How do you solve 3x^2 -12x = 9 using the quadratic formula?

Apr 24, 2017

$x = 2 \pm \sqrt{7}$

#### Explanation:

$\text{rearrange terms and equate to zero}$

$\Rightarrow 3 {x}^{2} - 12 x - 9 = 0$

$\text{divide all terms by 3}$

$\Rightarrow {x}^{2} - 4 x - 3 = 0$

$\text{compare to standard form } a {x}^{2} + b x + c = 0$

$\text{here " a=1,b=-4" and } c = - 3$

$\text{using the " color(blue)"quadratic formula}$

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{x = \frac{- b \pm \sqrt{\left({b}^{2} - 4 a c\right)}}{2 a}} \textcolor{w h i t e}{\frac{2}{2}} |}}}$

$\Rightarrow x = \frac{- \left(- 4\right) \pm \sqrt{{\left(- 4\right)}^{2} - \left(4 \times 1 \times - 3\right)}}{2 \times 1}$

$\textcolor{w h i t e}{\Rightarrow x} = \frac{4 \pm \sqrt{16 + 12}}{2} = \frac{4 \pm \sqrt{28}}{2}$

$\textcolor{w h i t e}{\Rightarrow x} = 2 \pm \frac{1}{2} \sqrt{28}$

$\textcolor{w h i t e}{\Rightarrow x} = 2 \pm \frac{1}{2} \sqrt{4 \times 7}$

$\textcolor{w h i t e}{\Rightarrow x} = 2 \pm \sqrt{7}$