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4n^2+12n-91=0
a = 4; b = 12; c = -91;
Δ = b2-4ac
Δ = 122-4·4·(-91)
Δ = 1600
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{1600}=40$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-40}{2*4}=\frac{-52}{8} =-6+1/2 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+40}{2*4}=\frac{28}{8} =3+1/2 $
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