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n^2-9n-52=0
a = 1; b = -9; c = -52;
Δ = b2-4ac
Δ = -92-4·1·(-52)
Δ = 289
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{289}=17$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-17}{2*1}=\frac{-8}{2} =-4 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+17}{2*1}=\frac{26}{2} =13 $
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