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2n^2+14n-104=0
a = 2; b = 14; c = -104;
Δ = b2-4ac
Δ = 142-4·2·(-104)
Δ = 1028
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}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1028}=\sqrt{4*257}=\sqrt{4}*\sqrt{257}=2\sqrt{257}$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-2\sqrt{257}}{2*2}=\frac{-14-2\sqrt{257}}{4} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+2\sqrt{257}}{2*2}=\frac{-14+2\sqrt{257}}{4} $
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