-p2+10p+100=0

Solution for -p2+10p+100=0 equation:

-p2+10p+100=0

We add all the numbers together, and all the variables

-1p^2+10p+100=0

a = -1; b = 10; c = +100;
Δ = b2-4ac
Δ = 102-4·(-1)·100
Δ = 500
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:

$\sqrt{\Delta}=\sqrt{500}=\sqrt{100*5}=\sqrt{100}*\sqrt{5}=10\sqrt{5}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-10\sqrt{5}}{2*-1}=\frac{-10-10\sqrt{5}}{-2}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+10\sqrt{5}}{2*-1}=\frac{-10+10\sqrt{5}}{-2}
$