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-2n^2+3n+5=0
a = -2; b = 3; c = +5;
Δ = b2-4ac
Δ = 32-4·(-2)·5
Δ = 49
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{49}=7$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-7}{2*-2}=\frac{-10}{-4} =2+1/2 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+7}{2*-2}=\frac{4}{-4} =-1 $
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