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2n^2-30=0
a = 2; b = 0; c = -30;
Δ = b2-4ac
Δ = 02-4·2·(-30)
Δ = 240
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}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{240}=\sqrt{16*15}=\sqrt{16}*\sqrt{15}=4\sqrt{15}$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{15}}{2*2}=\frac{0-4\sqrt{15}}{4} =-\frac{4\sqrt{15}}{4} =-\sqrt{15} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{15}}{2*2}=\frac{0+4\sqrt{15}}{4} =\frac{4\sqrt{15}}{4} =\sqrt{15} $
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