-3p^2-12p+8=0

Solution for -3p^2-12p+8=0 equation:

-3p^2-12p+8=0

a = -3; b = -12; c = +8;
Δ = b2-4ac
Δ = -122-4·(-3)·8
Δ = 240
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:

$\sqrt{\Delta}=\sqrt{240}=\sqrt{16*15}=\sqrt{16}*\sqrt{15}=4\sqrt{15}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4\sqrt{15}}{2*-3}=\frac{12-4\sqrt{15}}{-6}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4\sqrt{15}}{2*-3}=\frac{12+4\sqrt{15}}{-6}
$