# How do you factor and solve 6x^2 - 5x = 6?

Aug 21, 2016

#### Answer:

$x = - \frac{2}{3}$ or $x = \frac{3}{2}$

#### Explanation:

First subtract $6$ from both sides to get:

$6 {x}^{2} - 5 x - 6 = 0$

Factor using an AC method:

Find a pair of factors of $A C = 6 \cdot 6 = 36$ which differ by $B = 5$.

The pair $9 , 4$ works. Use this pair to split the middle term, then factor by grouping:

$0 = 6 {x}^{2} - 5 x - 6$

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

$= \left(6 {x}^{2} - 9 x\right) + \left(4 x - 6\right)$

$= 3 x \left(2 x - 3\right) + 2 \left(2 x - 3\right)$

$= \left(3 x + 2\right) \left(2 x - 3\right)$

Hence zeros: $x = - \frac{2}{3}$ and $x = \frac{3}{2}$

Aug 21, 2016

#### Answer:

The solutions are the values of $x$ that satisfy $6 {x}^{2} - 5 x - 6 = 0$

#### Explanation:

By completing the square you can get the formula for the solutions (roots) of a 2nd degree polynomial equation $a {x}^{2} + b x + c = 0$, as follows ${x}_{1} {,}_{2} = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$. In this case we have $a = 6$, $b = - 5$ and $c = - 6$

So the first solution is ${x}_{1} = \frac{- b + \sqrt{{b}^{2} - 4 a c}}{2 a} = \frac{5 + \sqrt{25 - 4 \cdot 6 \cdot \left(- 6\right)}}{2 \cdot 6} = \frac{5 + \sqrt{25 + 144}}{12} = \frac{5 + \sqrt{169}}{12} = \frac{5 + 13}{12} = \frac{18}{12} = \frac{3}{2}$

Similarly, the second solution is $\frac{5 - 13}{12} = - \frac{8}{12} = - \frac{2}{3}$