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3n^2=45
We move all terms to the left:
3n^2-(45)=0
a = 3; b = 0; c = -45;
Δ = b2-4ac
Δ = 02-4·3·(-45)
Δ = 540
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}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{540}=\sqrt{36*15}=\sqrt{36}*\sqrt{15}=6\sqrt{15}$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{15}}{2*3}=\frac{0-6\sqrt{15}}{6} =-\frac{6\sqrt{15}}{6} =-\sqrt{15} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{15}}{2*3}=\frac{0+6\sqrt{15}}{6} =\frac{6\sqrt{15}}{6} =\sqrt{15} $
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