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n^2=4590
We move all terms to the left:
n^2-(4590)=0
a = 1; b = 0; c = -4590;
Δ = b2-4ac
Δ = 02-4·1·(-4590)
Δ = 18360
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}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{18360}=\sqrt{36*510}=\sqrt{36}*\sqrt{510}=6\sqrt{510}$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{510}}{2*1}=\frac{0-6\sqrt{510}}{2} =-\frac{6\sqrt{510}}{2} =-3\sqrt{510} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{510}}{2*1}=\frac{0+6\sqrt{510}}{2} =\frac{6\sqrt{510}}{2} =3\sqrt{510} $
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