Question

What are the x-intercepts of 4x^ + 8x - 5

Answer 1

Assuming the exponent is supposed to be "^2" your equation will read:

4x² + 8x - 5

It must set equal to y to be a valid function and the y must be set equal to zero to find x-intercepts.

4x² + 8x - 5 = 0
4x² - 2x + 10x - 5 = 0
2x(2x - 1) + 5(2x - 1) = 0

(2x + 5)(2x - 1) = 0

Set each binomial equal to zero.

2x + 5 = 0
2x = 0 - 5
2x = - 5
Divide both sides by 2
x = - 5/2

2x - 1 = 0
2x = 0 + 1
2x = 1
x = 1/2

Your x-intercepts are x = - 5/2, 1/2 or (- 5/2, 0) and (1/2, 0)