# How many real roots are in y=2x^2-7x-15?

Mar 29, 2018

The real roots in $y = 2 {x}^{2} - 7 x - 15$ are $x = - \frac{3}{2}$ and $x = 5$.

#### Explanation:

$y = 2 {x}^{2} - 7 x - 15$

$y = 0$

$2 {x}^{2} - 7 x - 15 = 0$

Attempting to factorise

$2 {x}^{2} + 3 x - 10 x - 15 = 0$

$x \left(2 x + 3\right) - 5 \left(2 x + 3\right) = 0$

$\left(x - 5\right) \left(2 x + 3\right) = 0$

$x - 5 = 0$

$x = 5$

$2 x + 3 = 0$

$2 x = - 3$

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

The real roots in $y = 2 {x}^{2} - 7 x - 15$ are $x = - \frac{3}{2}$ and $x = 5$.