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2n^2-11n-21=0
a = 2; b = -11; c = -21;
Δ = b2-4ac
Δ = -112-4·2·(-21)
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$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-17}{2*2}=\frac{-6}{4} =-1+1/2 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+17}{2*2}=\frac{28}{4} =7 $
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