Question

What are the zeros of the quadratic function?
y = x^2 - 2x - 15

Answer 1

Answer:

$x=\frac{1+\sqrt{31}}{2},\:x=\frac{1-\sqrt{31}}{2}$

Step-by-step explanation:

$-> 0=2x^2-2x-15$

-zeros means the solution aka x intercept, meaning y=0

$x_{1,\:2}=\frac{-\left(-2\right)\pm \sqrt{\left(-2\right)^2-4\cdot \:2\left(-15\right)}}{2\cdot \:2}$

=> according to quadratic formula

$x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

where $ax^2+bx+c=0$

=> $x_{1,\:2}=\frac{-\left(-2\right)\pm \:2\sqrt{31}}{2\cdot \:2}$

simplify

$x_1=\frac{-\left(-2\right)+2\sqrt{31}}{2\cdot \:2},\:x_2=\frac{-\left(-2\right)-2\sqrt{31}}{2\cdot \:2}$

$x=\frac{1+\sqrt{31}}{2},\:x=\frac{1-\sqrt{31}}{2}$

Cannot simplify farther