p^2+2p-135=0

Solution for p^2+2p-135=0 equation:

p^2+2p-135=0

a = 1; b = 2; c = -135;
Δ = b2-4ac
Δ = 22-4·1·(-135)
Δ = 544
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:

$\sqrt{\Delta}=\sqrt{544}=\sqrt{16*34}=\sqrt{16}*\sqrt{34}=4\sqrt{34}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-4\sqrt{34}}{2*1}=\frac{-2-4\sqrt{34}}{2}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+4\sqrt{34}}{2*1}=\frac{-2+4\sqrt{34}}{2}
$