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2n^2-130=0
a = 2; b = 0; c = -130;
Δ = b2-4ac
Δ = 02-4·2·(-130)
Δ = 1040
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}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1040}=\sqrt{16*65}=\sqrt{16}*\sqrt{65}=4\sqrt{65}$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{65}}{2*2}=\frac{0-4\sqrt{65}}{4} =-\frac{4\sqrt{65}}{4} =-\sqrt{65} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{65}}{2*2}=\frac{0+4\sqrt{65}}{4} =\frac{4\sqrt{65}}{4} =\sqrt{65} $
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