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n^2+16n-2=6
We move all terms to the left:
n^2+16n-2-(6)=0
We add all the numbers together, and all the variables
n^2+16n-8=0
a = 1; b = 16; c = -8;
Δ = b2-4ac
Δ = 162-4·1·(-8)
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{-(16)-12\sqrt{2}}{2*1}=\frac{-16-12\sqrt{2}}{2} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+12\sqrt{2}}{2*1}=\frac{-16+12\sqrt{2}}{2} $
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