# How do you find the zeros, real and imaginary, of y=- 5x^2+15x+10  using the quadratic formula?

Feb 17, 2016

The answers are: $x = \frac{3 \pm \sqrt{17}}{2}$, which are approximately $- 0.56$ and $3.56$.

#### Explanation:

The quadratic formula says the zeros of $y = a {x}^{2} + b x + c$ are $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$.

For this example, $a = - 5$, $b = 15$, and $c = 10$, so we get:

x=(-15 pm sqrt( (15)^2-4 * (-5) * 10) )/(2*(-5)) =(15 pm sqrt(225+200))/10

$= \frac{15 \pm \sqrt{425}}{10} = \frac{15 \pm \sqrt{25} \sqrt{17}}{10}$

$= \frac{15 \pm 5 \sqrt{17}}{10} = \frac{3 \pm \sqrt{17}}{2}$