# How do you solve 3x^2-12x+11=0 using the quadratic function?

Jun 27, 2018

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

#### Explanation:

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

for a quadratic taking the form

$a {x}^{2} + b x + c = 0$

So for this equation it is

$x = \frac{- \left(- 12\right) \pm \sqrt{{\left(- 12\right)}^{2} - 4 \cdot 3 \cdot 11}}{2 \cdot 3}$

$x = \frac{12 \pm \sqrt{144 - 132}}{6}$

$x = \frac{12 \pm \sqrt{4 \cdot 3}}{6}$

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

Jul 12, 2018

color(blue)(x = 2 + 1/sqrt3, 2 - 1/sqrt3= 2.5774, 1.4226

#### Explanation:

Quadratic formula to find the roots $x = \frac{- b \pm \sqrt{D}}{2 a}$ where D is the discrimination to decide whether the roots are real or imaginary.

$D = {b}^{2} - 4 a c$

Given equation is $3 {x}^{2} - 12 x + 11 = 0$

$a = 3 , b = - 12 , c = 11 , D = - {12}^{2} - \left(4 \cdot 3 \cdot 11\right) = 12$

$x = \frac{12 \pm \sqrt{12}}{6}$

$x = 2 \pm \left(\frac{1}{\sqrt{3}}\right)$

color(blue)(x = 2 + 1/sqrt3, 2 - 1/sqrt3= 2.5774, 1.4226