-p^2+10p+5=0

Solution for -p^2+10p+5=0 equation:

-p^2+10p+5=0

We add all the numbers together, and all the variables

-1p^2+10p+5=0

a = -1; b = 10; c = +5;
Δ = b2-4ac
Δ = 102-4·(-1)·5
Δ = 120
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:

$\sqrt{\Delta}=\sqrt{120}=\sqrt{4*30}=\sqrt{4}*\sqrt{30}=2\sqrt{30}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{30}}{2*-1}=\frac{-10-2\sqrt{30}}{-2}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{30}}{2*-1}=\frac{-10+2\sqrt{30}}{-2}
$