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2n^2+2n-289=0
a = 2; b = 2; c = -289;
Δ = b2-4ac
Δ = 22-4·2·(-289)
Δ = 2316
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}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2316}=\sqrt{4*579}=\sqrt{4}*\sqrt{579}=2\sqrt{579}$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{579}}{2*2}=\frac{-2-2\sqrt{579}}{4} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{579}}{2*2}=\frac{-2+2\sqrt{579}}{4} $
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