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p^2+8p-72=-8
We move all terms to the left:
p^2+8p-72-(-8)=0
We add all the numbers together, and all the variables
p^2+8p-64=0
a = 1; b = 8; c = -64;
Δ = b2-4ac
Δ = 82-4·1·(-64)
Δ = 320
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}$$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{320}=\sqrt{64*5}=\sqrt{64}*\sqrt{5}=8\sqrt{5}$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-8\sqrt{5}}{2*1}=\frac{-8-8\sqrt{5}}{2} $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+8\sqrt{5}}{2*1}=\frac{-8+8\sqrt{5}}{2} $
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