-4x^2-2x+36=0

Solution for -4x^2-2x+36=0 equation:

-4x^2-2x+36=0

a = -4; b = -2; c = +36;
Δ = b2-4ac
Δ = -22-4·(-4)·36
Δ = 580
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{580}=\sqrt{4*145}=\sqrt{4}*\sqrt{145}=2\sqrt{145}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{145}}{2*-4}=\frac{2-2\sqrt{145}}{-8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{145}}{2*-4}=\frac{2+2\sqrt{145}}{-8}
$