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27a^2+a-12=0
a = 27; b = 1; c = -12;
Δ = b2-4ac
Δ = 12-4·27·(-12)
Δ = 1297
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{1297}}{2*27}=\frac{-1-\sqrt{1297}}{54} $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{1297}}{2*27}=\frac{-1+\sqrt{1297}}{54} $
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