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-3x^2+x+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:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{145}}{2*-3}=\frac{-1-\sqrt{145}}{-6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{145}}{2*-3}=\frac{-1+\sqrt{145}}{-6} $
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