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6x^2+-19x+-10=0
We add all the numbers together, and all the variables
6x^2-19x=0
a = 6; b = -19; c = 0;
Δ = b2-4ac
Δ = -192-4·6·0
Δ = 361
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{361}=19$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-19}{2*6}=\frac{0}{12} =0 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+19}{2*6}=\frac{38}{12} =3+1/6 $
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