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2n^2+15n-108=0
a = 2; b = 15; c = -108;
Δ = b2-4ac
Δ = 152-4·2·(-108)
Δ = 1089
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{1089}=33$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-33}{2*2}=\frac{-48}{4} =-12 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+33}{2*2}=\frac{18}{4} =4+1/2 $
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