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2n^2-2n-12=0
a = 2; b = -2; c = -12;
Δ = b2-4ac
Δ = -22-4·2·(-12)
Δ = 100
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{100}=10$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-10}{2*2}=\frac{-8}{4} =-2 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+10}{2*2}=\frac{12}{4} =3 $
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