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n^2+18n=13
We move all terms to the left:
n^2+18n-(13)=0
a = 1; b = 18; c = -13;
Δ = b2-4ac
Δ = 182-4·1·(-13)
Δ = 376
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}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{376}=\sqrt{4*94}=\sqrt{4}*\sqrt{94}=2\sqrt{94}$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-2\sqrt{94}}{2*1}=\frac{-18-2\sqrt{94}}{2} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+2\sqrt{94}}{2*1}=\frac{-18+2\sqrt{94}}{2} $
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