p^2=-10p

Solution for p^2=-10p equation:

p^2=-10p

We move all terms to the left:

p^2-(-10p)=0

We get rid of parentheses

p^2+10p=0

a = 1; b = 10; c = 0;
Δ = b2-4ac
Δ = 102-4·1·0
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-10}{2*1}=\frac{-20}{2}
=-10
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+10}{2*1}=\frac{0}{2}
=0
$