-3p^2+20p+24=0

Solution for -3p^2+20p+24=0 equation:

-3p^2+20p+24=0

a = -3; b = 20; c = +24;
Δ = b2-4ac
Δ = 202-4·(-3)·24
Δ = 688
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:

$\sqrt{\Delta}=\sqrt{688}=\sqrt{16*43}=\sqrt{16}*\sqrt{43}=4\sqrt{43}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{43}}{2*-3}=\frac{-20-4\sqrt{43}}{-6}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{43}}{2*-3}=\frac{-20+4\sqrt{43}}{-6}
$