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6x^2-19x-7=0
a = 6; b = -19; c = -7;
Δ = b2-4ac
Δ = -192-4·6·(-7)
Δ = 529
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{529}=23$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-23}{2*6}=\frac{-4}{12} =-1/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+23}{2*6}=\frac{42}{12} =3+1/2 $
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