-2p^2+3p+36=0

Solution for -2p^2+3p+36=0 equation:

-2p^2+3p+36=0

a = -2; b = 3; c = +36;
Δ = b2-4ac
Δ = 32-4·(-2)·36
Δ = 297
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{297}=\sqrt{9*33}=\sqrt{9}*\sqrt{33}=3\sqrt{33}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3\sqrt{33}}{2*-2}=\frac{-3-3\sqrt{33}}{-4}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3\sqrt{33}}{2*-2}=\frac{-3+3\sqrt{33}}{-4}
$