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(6x^2)+19x+14=0
a = 6; b = 19; c = +14;
Δ = b2-4ac
Δ = 192-4·6·14
Δ = 25
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{25}=5$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-5}{2*6}=\frac{-24}{12} =-2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+5}{2*6}=\frac{-14}{12} =-1+1/6 $
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