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(12n^2)-3n-5=0
a = 12; b = -3; c = -5;
Δ = b2-4ac
Δ = -32-4·12·(-5)
Δ = 249
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}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-\sqrt{249}}{2*12}=\frac{3-\sqrt{249}}{24} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{249}}{2*12}=\frac{3+\sqrt{249}}{24} $
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