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2n^2=240
We move all terms to the left:
2n^2-(240)=0
a = 2; b = 0; c = -240;
Δ = b2-4ac
Δ = 02-4·2·(-240)
Δ = 1920
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}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1920}=\sqrt{64*30}=\sqrt{64}*\sqrt{30}=8\sqrt{30}$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{30}}{2*2}=\frac{0-8\sqrt{30}}{4} =-\frac{8\sqrt{30}}{4} =-2\sqrt{30} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{30}}{2*2}=\frac{0+8\sqrt{30}}{4} =\frac{8\sqrt{30}}{4} =2\sqrt{30} $
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