4p^2+4p-5=0

Solution for 4p^2+4p-5=0 equation:

4p^2+4p-5=0

a = 4; b = 4; c = -5;
Δ = b2-4ac
Δ = 42-4·4·(-5)
Δ = 96
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:

$\sqrt{\Delta}=\sqrt{96}=\sqrt{16*6}=\sqrt{16}*\sqrt{6}=4\sqrt{6}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{6}}{2*4}=\frac{-4-4\sqrt{6}}{8}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{6}}{2*4}=\frac{-4+4\sqrt{6}}{8}
$