p^2+10-56=0

Solution for p^2+10-56=0 equation:

p^2+10-56=0

We add all the numbers together, and all the variables

p^2-46=0

a = 1; b = 0; c = -46;
Δ = b2-4ac
Δ = 02-4·1·(-46)
Δ = 184
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:

$\sqrt{\Delta}=\sqrt{184}=\sqrt{4*46}=\sqrt{4}*\sqrt{46}=2\sqrt{46}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{46}}{2*1}=\frac{0-2\sqrt{46}}{2}
=-\frac{2\sqrt{46}}{2}
=-\sqrt{46}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{46}}{2*1}=\frac{0+2\sqrt{46}}{2}
=\frac{2\sqrt{46}}{2}
=\sqrt{46}
$