# How do you solve 3x^2 + 18x - 12 = 0?

Nov 26, 2015

The solutions are
color(blue)(x =(-9+sqrt(117))/3

color(blue)(x =(-9-sqrt(117))/3

#### Explanation:

$3 {x}^{2} + 18 x - 12 = 0$

The equation is of the form color(blue)(ax^2+bx+c=0 where:
$a = 3 , b = 18 , c = - 12$

The Discriminant is given by:
color(blue)(Delta=b^2-4*a*c

$= {\left(18\right)}^{2} - \left(4 \cdot 3 \cdot \left(- 12\right)\right)$

$= 324 + 144 = 468$

The solutions are found using the formula:
color(blue)(x=(-b+-sqrtDelta)/(2*a)

$x = \frac{\left(- 18\right) \pm \sqrt{468}}{2 \cdot 3} = \frac{- 18 \pm \sqrt{468}}{6}$

(note:$\sqrt{468} = \sqrt{2 \cdot 2 \cdot 117} = 2 \sqrt{117}$)

$x = \frac{- 18 \pm 2 \sqrt{117}}{6}$

$x = \frac{\cancel{2} \cdot \left(- 9 \pm \sqrt{117}\right)}{\cancel{6}}$

$x = \frac{- 9 \pm \sqrt{117}}{3}$

The solutions are
color(blue)(x =(-9+sqrt(117))/3

color(blue)(x =(-9-sqrt(117))/3

Nov 26, 2015

$x = - 3 + \sqrt{13}$

$x = - 3 - \sqrt{13}$

#### Explanation:

$3 {x}^{2} + 18 x - 12 = 0$

Divide both sides by $3$.

${x}^{2} + 6 x - 4 = 0$ is a quadratic equation, $a {x}^{2} + b x + c$, where

$a = 1 , b = 6 , c = - 4$

Solve by completing the square.

Add $4$ to both sides of the equation.

${x}^{2} + 6 x = 4$

Divide $b$ by $2$ and square the result. Add to both sides of the equation.

${\left(\frac{6}{2}\right)}^{2} = 9$

${x}^{2} + 6 x + 9 = 4 + 9$

${x}^{2} + 6 x + 9 = 13$

Write the trinomial as a perfect square.

${\left(x + 3\right)}^{2} = 13$

Take the square root of both sides.

$x + 3 = \pm \sqrt{13}$

Subtract $3$ from both sides.

$x = - 3 \pm \sqrt{13}$

Solve for $x$.

$x = - 3 + \sqrt{13}$

$x = - 3 - \sqrt{13}$