p^2=4p+28

Solution for p^2=4p+28 equation:

p^2=4p+28

We move all terms to the left:

p^2-(4p+28)=0

We get rid of parentheses

p^2-4p-28=0

a = 1; b = -4; c = -28;
Δ = b2-4ac
Δ = -42-4·1·(-28)
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{-(-4)-8\sqrt{2}}{2*1}=\frac{4-8\sqrt{2}}{2}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+8\sqrt{2}}{2*1}=\frac{4+8\sqrt{2}}{2}
$