p^2+10p-36=0

Solution for p^2+10p-36=0 equation:

p^2+10p-36=0

a = 1; b = 10; c = -36;
Δ = b2-4ac
Δ = 102-4·1·(-36)
Δ = 244
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:

$\sqrt{\Delta}=\sqrt{244}=\sqrt{4*61}=\sqrt{4}*\sqrt{61}=2\sqrt{61}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{61}}{2*1}=\frac{-10-2\sqrt{61}}{2}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{61}}{2*1}=\frac{-10+2\sqrt{61}}{2}
$