p^2=4p+4

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



p^2=4p+4
We move all terms to the left:
p^2-(4p+4)=0
We get rid of parentheses
p^2-4p-4=0
a = 1; b = -4; c = -4;
Δ = b2-4ac
Δ = -42-4·1·(-4)
Δ = 32
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:
$\sqrt{\Delta}=\sqrt{32}=\sqrt{16*2}=\sqrt{16}*\sqrt{2}=4\sqrt{2}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4\sqrt{2}}{2*1}=\frac{4-4\sqrt{2}}{2} $
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4\sqrt{2}}{2*1}=\frac{4+4\sqrt{2}}{2} $

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