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n^2=58
We move all terms to the left:
n^2-(58)=0
a = 1; b = 0; c = -58;
Δ = b2-4ac
Δ = 02-4·1·(-58)
Δ = 232
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}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{232}=\sqrt{4*58}=\sqrt{4}*\sqrt{58}=2\sqrt{58}$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{58}}{2*1}=\frac{0-2\sqrt{58}}{2} =-\frac{2\sqrt{58}}{2} =-\sqrt{58} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{58}}{2*1}=\frac{0+2\sqrt{58}}{2} =\frac{2\sqrt{58}}{2} =\sqrt{58} $
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