Question

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

Answer 1

`x=(1+√(31))/(2),\:x=(1-√(31))/(2)`

Explanation:

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

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

`x_(1,\:2)=(-\left(-2\right)\pm √(\left(-2\right)^2-4\cdot \:2\left(-15\right)))/(2\cdot \:2)`

=> according to quadratic formula

`x_(1,\:2)=(-b\pm √(b^2-4ac))/(2a)`

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

=>
`x_(1,\:2)=(-\left(-2\right)\pm \:2√(31))/(2\cdot \:2)`

simplify

`x_1=(-\left(-2\right)+2√(31))/(2\cdot \:2),\:x_2=(-\left(-2\right)-2√(31))/(2\cdot \:2)`

`x=(1+√(31))/(2),\:x=(1-√(31))/(2)`

Cannot simplify farther