How do you solve 9x^2 - 7x = 12 using completing the square?

Jun 19, 2015

I found:
${x}_{1} = \frac{1}{18} \left(7 + \sqrt{481}\right)$
${x}_{2} = \frac{1}{18} \left(7 - \sqrt{481}\right)$

Explanation:

Let us do some manipulations:

$9 {x}^{2} - 7 x = 12$ add and subtract $\frac{49}{36}$
$9 {x}^{2} - 7 x + \frac{49}{36} - \frac{49}{36} = 12$
$9 {x}^{2} - 7 x + \frac{49}{36} = 12 + \frac{49}{36}$
${\left(3 x - \frac{7}{6}\right)}^{2} = \frac{481}{36}$ root square both sides:
$3 x - \frac{7}{6} = \pm \sqrt{\frac{481}{36}} =$
$x = \frac{1}{3} \left(\frac{7}{6} \pm \sqrt{\frac{481}{36}}\right)$
${x}_{1} = \frac{1}{18} \left(7 + \sqrt{481}\right)$
${x}_{2} = \frac{1}{18} \left(7 - \sqrt{481}\right)$