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12n^2+9n-12=0
a = 12; b = 9; c = -12;
Δ = b2-4ac
Δ = 92-4·12·(-12)
Δ = 657
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}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{657}=\sqrt{9*73}=\sqrt{9}*\sqrt{73}=3\sqrt{73}$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-3\sqrt{73}}{2*12}=\frac{-9-3\sqrt{73}}{24} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+3\sqrt{73}}{2*12}=\frac{-9+3\sqrt{73}}{24} $
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