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3n^2=162
We move all terms to the left:
3n^2-(162)=0
a = 3; b = 0; c = -162;
Δ = b2-4ac
Δ = 02-4·3·(-162)
Δ = 1944
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}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1944}=\sqrt{324*6}=\sqrt{324}*\sqrt{6}=18\sqrt{6}$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-18\sqrt{6}}{2*3}=\frac{0-18\sqrt{6}}{6} =-\frac{18\sqrt{6}}{6} =-3\sqrt{6} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+18\sqrt{6}}{2*3}=\frac{0+18\sqrt{6}}{6} =\frac{18\sqrt{6}}{6} =3\sqrt{6} $
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