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3n^2-n-12=0
We add all the numbers together, and all the variables
3n^2-1n-12=0
a = 3; b = -1; c = -12;
Δ = b2-4ac
Δ = -12-4·3·(-12)
Δ = 145
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{-(-1)-\sqrt{145}}{2*3}=\frac{1-\sqrt{145}}{6} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{145}}{2*3}=\frac{1+\sqrt{145}}{6} $
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