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n^2-25n+45=0
a = 1; b = -25; c = +45;
Δ = b2-4ac
Δ = -252-4·1·45
Δ = 445
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{445}}{2*1}=\frac{25-\sqrt{445}}{2} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+\sqrt{445}}{2*1}=\frac{25+\sqrt{445}}{2} $
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