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6a^2-19a+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:$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{25}=5$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-5}{2*6}=\frac{14}{12} =1+1/6 $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+5}{2*6}=\frac{24}{12} =2 $
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