-16x^2+36x+4=0

Solution for -16x^2+36x+4=0 equation:

-16x^2+36x+4=0

a = -16; b = 36; c = +4;
Δ = b2-4ac
Δ = 362-4·(-16)·4
Δ = 1552
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:

$\sqrt{\Delta}=\sqrt{1552}=\sqrt{16*97}=\sqrt{16}*\sqrt{97}=4\sqrt{97}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(36)-4\sqrt{97}}{2*-16}=\frac{-36-4\sqrt{97}}{-32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(36)+4\sqrt{97}}{2*-16}=\frac{-36+4\sqrt{97}}{-32}
$