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6x^2+19x+12=0
a = 6; b = 19; c = +12;
Δ = b2-4ac
Δ = 192-4·6·12
Δ = 73
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{-(19)-\sqrt{73}}{2*6}=\frac{-19-\sqrt{73}}{12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+\sqrt{73}}{2*6}=\frac{-19+\sqrt{73}}{12} $
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