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14n^2+50n-24=0
a = 14; b = 50; c = -24;
Δ = b2-4ac
Δ = 502-4·14·(-24)
Δ = 3844
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}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{3844}=62$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(50)-62}{2*14}=\frac{-112}{28} =-4 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(50)+62}{2*14}=\frac{12}{28} =3/7 $
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