# How do you factor 2x^2+5x-6?

Oct 10, 2015

$2 {x}^{2} + 5 x - 6 = 2 \left(x + \frac{5}{4} - \frac{\sqrt{73}}{4}\right) \left(x + \frac{5}{4} + \frac{\sqrt{73}}{4}\right)$

#### Explanation:

Quadratic formula states that for an equation of the form $a {x}^{2} + b x + c = 0$,

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

These two numbers are called roots.

Since plugging these 2 roots to $a {x}^{2} + b x + c$ gives 0, we can say that (x-root1)(x-root2)=0.

Thus, finding these roots means factoring the equation!
In this problem, $a = 2 , b = 5 , \mathmr{and} c = - 6$.

So

${x}_{1} = \frac{- 5 + \sqrt{{5}^{2} - 4 \left(2\right) \left(- 6\right)}}{2 \left(2\right)}$, ${x}_{2} = \frac{- 5 - \sqrt{{5}^{2} - 4 \left(2\right) \left(- 6\right)}}{2 \left(2\right)}$.

${x}_{1} = \frac{- 5 + \sqrt{73}}{4}$, ${x}_{2} = \frac{- 5 - \sqrt{73}}{4}$.

In conclusion,

$2 {x}^{2} + 5 x - 6 = 2 \left(x + \frac{5}{4} - \frac{\sqrt{73}}{4}\right) \left(x + \frac{5}{4} + \frac{\sqrt{73}}{4}\right)$.

I multiplied the whole factored equation by 2 because the factored equation starts with ${x}^{2}$ and the original equation has $2 {x}^{2}$.