n2+8=80

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Solution for n2+8=80 equation:



n2+8=80
We move all terms to the left:
n2+8-(80)=0
We add all the numbers together, and all the variables
n^2-72=0
a = 1; b = 0; c = -72;
Δ = b2-4ac
Δ = 02-4·1·(-72)
Δ = 288
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:
$\sqrt{\Delta}=\sqrt{288}=\sqrt{144*2}=\sqrt{144}*\sqrt{2}=12\sqrt{2}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12\sqrt{2}}{2*1}=\frac{0-12\sqrt{2}}{2} =-\frac{12\sqrt{2}}{2} =-6\sqrt{2} $
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12\sqrt{2}}{2*1}=\frac{0+12\sqrt{2}}{2} =\frac{12\sqrt{2}}{2} =6\sqrt{2} $

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