# How do you solve using the quadratic formula 6x+9=2x^2?

Apr 13, 2018

$x = \frac{3}{2} + \frac{3}{2} \sqrt{3}$ or $x = \frac{3}{2} - \frac{3}{2} \sqrt{3}$

#### Explanation:

$6 x + 9 = 2 {x}^{2}$

$\therefore 2 {x}^{2} = 6 x + 9$

$\therefore 2 {x}^{2} - 6 x - 9 = 0$

Quadratic formula:- $x = a {x}^{2} + b x + c$

$\therefore a = 2 , b = - 6 \mathmr{and} c = - 9$

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

$\therefore x = \frac{- \left(- 6\right) \pm \sqrt{\left(- {6}^{2}\right) - 4 \left(2\right) \left(- 9\right)}}{2 \left(2\right)}$

$\therefore x = \frac{6 \pm \sqrt{108}}{4}$

$\therefore x = \frac{6 \pm \sqrt{3 \cdot 3 \cdot 3 \cdot 2 \cdot 2}}{4}$

$\therefore \sqrt{2} \cdot \sqrt{2} = 2$

$\therefore \sqrt{3} \cdot \sqrt{3} = 3$

$\therefore x = \frac{6 \pm 3 \cdot 2 \sqrt{3}}{4}$

$\therefore x = \frac{6 \pm 6 \sqrt{3}}{4}$

$\therefore x = {\cancel{6}}^{3} / {\cancel{4}}^{2} \pm \frac{{\cancel{6}}^{3} \sqrt{3}}{\cancel{4}} ^ 2$

$\therefore x = \frac{3}{2} \pm \frac{3 \sqrt{3}}{2}$

$\therefore x = \frac{3}{2} + \frac{3}{2} \sqrt{3}$ or $x = \frac{3}{2} - \frac{3}{2} \sqrt{3}$

Apr 13, 2018

x=3/2+−3/2sqrt(3) or x=3/2+3/2sqrt3

#### Explanation:

Step 1: Subtract $2 {x}^{2}$ from both sides.
6x+9−color(blue)(2x^2)=2x^2−color(blue)(2x^2)
−2x^2+6x+9=0
Step 2: Use quadratic formula with $\textcolor{red}{a = - 2} , \textcolor{p u r p \le}{b = 6} , \textcolor{g r e e n}{c = 9} .$
x=−b±sqrt(b^2−4ac)/(2a)
x=−(color(purple)(6))^2±sqrt((color(purple)(6))^2−4(color(red)(−2))(color(green)(9)))/(2(color(red)(−2))
x=
−6±sqrt108/(−4)
x=3/2+−3/2sqrt3 or x=3/2+3/2 sqrt3