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2n^2+4n-240=0
a = 2; b = 4; c = -240;
Δ = b2-4ac
Δ = 42-4·2·(-240)
Δ = 1936
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{1936}=44$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-44}{2*2}=\frac{-48}{4} =-12 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+44}{2*2}=\frac{40}{4} =10 $
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