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12n^2+12n-18=0
a = 12; b = 12; c = -18;
Δ = b2-4ac
Δ = 122-4·12·(-18)
Δ = 1008
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}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1008}=\sqrt{144*7}=\sqrt{144}*\sqrt{7}=12\sqrt{7}$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-12\sqrt{7}}{2*12}=\frac{-12-12\sqrt{7}}{24} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+12\sqrt{7}}{2*12}=\frac{-12+12\sqrt{7}}{24} $
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