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(2n^2)-25n-95=0
a = 2; b = -25; c = -95;
Δ = b2-4ac
Δ = -252-4·2·(-95)
Δ = 1385
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{-(-25)-\sqrt{1385}}{2*2}=\frac{25-\sqrt{1385}}{4} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+\sqrt{1385}}{2*2}=\frac{25+\sqrt{1385}}{4} $
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