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7n^2+23n-142=0
a = 7; b = 23; c = -142;
Δ = b2-4ac
Δ = 232-4·7·(-142)
Δ = 4505
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}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-\sqrt{4505}}{2*7}=\frac{-23-\sqrt{4505}}{14} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+\sqrt{4505}}{2*7}=\frac{-23+\sqrt{4505}}{14} $
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