-2p^2+22p+56=0

Solution for -2p^2+22p+56=0 equation:

-2p^2+22p+56=0

a = -2; b = 22; c = +56;
Δ = b2-4ac
Δ = 222-4·(-2)·56
Δ = 932
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:

$\sqrt{\Delta}=\sqrt{932}=\sqrt{4*233}=\sqrt{4}*\sqrt{233}=2\sqrt{233}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-2\sqrt{233}}{2*-2}=\frac{-22-2\sqrt{233}}{-4}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+2\sqrt{233}}{2*-2}=\frac{-22+2\sqrt{233}}{-4}
$