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n^2=110
We move all terms to the left:
n^2-(110)=0
a = 1; b = 0; c = -110;
Δ = b2-4ac
Δ = 02-4·1·(-110)
Δ = 440
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}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{440}=\sqrt{4*110}=\sqrt{4}*\sqrt{110}=2\sqrt{110}$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{110}}{2*1}=\frac{0-2\sqrt{110}}{2} =-\frac{2\sqrt{110}}{2} =-\sqrt{110} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{110}}{2*1}=\frac{0+2\sqrt{110}}{2} =\frac{2\sqrt{110}}{2} =\sqrt{110} $
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