# How do you solve 3x^2 + 17x -6 = 0 by completing the square?

Mar 3, 2018

color(blue)(x = 1/3, -6

#### Explanation:

$3 x 2 + 17 x - 6$ 0

$\cancel{3} \left({x}^{2} + \left(\frac{17}{3}\right) x - 2\right) = 0$

${x}^{2} + \textcolor{p u r p \le}{\left(2 \cdot \left(\frac{17}{6}\right) x\right)} = 2$ to make the middle term in the form of $2 x y$

Add color(green)( (17/6)^2 to both sides to make the L H S a perfect square.

x^2 + (2 * (17/6)x + (17/6)^2 = 2 + (17/6)^2 = 361/36

${\left(x + \left(\frac{17}{6}\right)\right)}^{2} = {\left(\frac{19}{6}\right)}^{2}$

$x + \frac{17}{6} = \textcolor{red}{\pm} \frac{19}{6}$

$x = \textcolor{red}{\pm} \frac{19}{6} - \frac{17}{6}$

color(blue)(x = 1/3, -6#