# How do you factor 2d^2 + 7d - 15?

Jun 4, 2015

$2 {d}^{2} + 7 d - 15 = \left(d + 5\right) \left(2 d - 3\right)$

I used the quadratic formula to solve this one.

$a = 2$, $b = 7$, $c = - 15$

$\Delta = {b}^{2} - 4 a c = 49 + 120 = 169 = {13}^{2}$

So $2 {d}^{2} + 7 d - 15 = 0$ has roots:

$d = \frac{- b \pm \sqrt{\Delta}}{2 a}$

$= \frac{- 7 \pm 13}{4}$

That is $d = - 5$ or $d = \frac{3}{2}$

Hence factors $\left(d + 5\right)$ and $\left(2 d - 3\right)$, respectively.