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4n^2+4n-35=0
a = 4; b = 4; c = -35;
Δ = b2-4ac
Δ = 42-4·4·(-35)
Δ = 576
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{576}=24$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-24}{2*4}=\frac{-28}{8} =-3+1/2 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+24}{2*4}=\frac{20}{8} =2+1/2 $
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