4p2+64p-88=0

Solution for 4p2+64p-88=0 equation:

4p^2+64p-88=0

a = 4; b = 64; c = -88;
Δ = b2-4ac
Δ = 642-4·4·(-88)
Δ = 5504
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:

$\sqrt{\Delta}=\sqrt{5504}=\sqrt{64*86}=\sqrt{64}*\sqrt{86}=8\sqrt{86}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(64)-8\sqrt{86}}{2*4}=\frac{-64-8\sqrt{86}}{8}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(64)+8\sqrt{86}}{2*4}=\frac{-64+8\sqrt{86}}{8}
$