p2-22p+56=0

Solution for p2-22p+56=0 equation:

p2-22p+56=0

We add all the numbers together, and all the variables

p^2-22p+56=0

a = 1; b = -22; c = +56;
Δ = b2-4ac
Δ = -222-4·1·56
Δ = 260
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:

$\sqrt{\Delta}=\sqrt{260}=\sqrt{4*65}=\sqrt{4}*\sqrt{65}=2\sqrt{65}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-2\sqrt{65}}{2*1}=\frac{22-2\sqrt{65}}{2}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+2\sqrt{65}}{2*1}=\frac{22+2\sqrt{65}}{2}
$