-3p^2+18p+96=0

Solution for -3p^2+18p+96=0 equation:

-3p^2+18p+96=0

a = -3; b = 18; c = +96;
Δ = b2-4ac
Δ = 182-4·(-3)·96
Δ = 1476
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:

$\sqrt{\Delta}=\sqrt{1476}=\sqrt{36*41}=\sqrt{36}*\sqrt{41}=6\sqrt{41}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-6\sqrt{41}}{2*-3}=\frac{-18-6\sqrt{41}}{-6}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+6\sqrt{41}}{2*-3}=\frac{-18+6\sqrt{41}}{-6}
$