# How do you solve using the completing the square method 3x^2-5x-6=0?

##### 1 Answer
May 10, 2016

$x = \frac{5}{6} \pm \frac{\sqrt{97}}{6}$

#### Explanation:

Use the difference of squares identity:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

with $a = \left(6 x - 5\right)$ and $b = \sqrt{97}$.

To avoid too much arithmetic with fractions, pre-multiply the equation by $3 \cdot {2}^{2} = 12$ first:

$0 = 12 \left(3 {x}^{2} - 5 x - 6\right)$

$= 36 {x}^{2} - 60 x - 72$

$= {\left(6 x - 5\right)}^{2} - {5}^{2} - 72$

$= {\left(6 x - 5\right)}^{2} - 25 - 72$

$= {\left(6 x - 5\right)}^{2} - 97$

$= {\left(6 x - 5\right)}^{2} - {\left(\sqrt{97}\right)}^{2}$

$= \left(\left(6 x - 5\right) - \sqrt{97}\right) \left(\left(6 x - 5\right) + \sqrt{97}\right)$

$= \left(6 x - 5 - \sqrt{97}\right) \left(6 x - 5 + \sqrt{97}\right)$

$= 36 \left(x - \frac{5}{6} - \frac{\sqrt{97}}{6}\right) \left(x - \frac{5}{6} + \frac{\sqrt{97}}{6}\right)$

So $x = \frac{5}{6} \pm \frac{\sqrt{97}}{6}$