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2n^2-10n-48=0
a = 2; b = -10; c = -48;
Δ = b2-4ac
Δ = -102-4·2·(-48)
Δ = 484
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{484}=22$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-22}{2*2}=\frac{-12}{4} =-3 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+22}{2*2}=\frac{32}{4} =8 $
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