p^2+15p-400=0

Solution for p^2+15p-400=0 equation:

p^2+15p-400=0

a = 1; b = 15; c = -400;
Δ = b2-4ac
Δ = 152-4·1·(-400)
Δ = 1825
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:

$\sqrt{\Delta}=\sqrt{1825}=\sqrt{25*73}=\sqrt{25}*\sqrt{73}=5\sqrt{73}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-5\sqrt{73}}{2*1}=\frac{-15-5\sqrt{73}}{2}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+5\sqrt{73}}{2*1}=\frac{-15+5\sqrt{73}}{2}
$