-p^2+10p+7=0

Solution for -p^2+10p+7=0 equation:

-p^2+10p+7=0

We add all the numbers together, and all the variables

-1p^2+10p+7=0

a = -1; b = 10; c = +7;
Δ = b2-4ac
Δ = 102-4·(-1)·7
Δ = 128
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:

$\sqrt{\Delta}=\sqrt{128}=\sqrt{64*2}=\sqrt{64}*\sqrt{2}=8\sqrt{2}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-8\sqrt{2}}{2*-1}=\frac{-10-8\sqrt{2}}{-2}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+8\sqrt{2}}{2*-1}=\frac{-10+8\sqrt{2}}{-2}
$