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6n^2=3
We move all terms to the left:
6n^2-(3)=0
a = 6; b = 0; c = -3;
Δ = b2-4ac
Δ = 02-4·6·(-3)
Δ = 72
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}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{72}=\sqrt{36*2}=\sqrt{36}*\sqrt{2}=6\sqrt{2}$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{2}}{2*6}=\frac{0-6\sqrt{2}}{12} =-\frac{6\sqrt{2}}{12} =-\frac{\sqrt{2}}{2} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{2}}{2*6}=\frac{0+6\sqrt{2}}{12} =\frac{6\sqrt{2}}{12} =\frac{\sqrt{2}}{2} $
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