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10x^2-19x+6=0
a = 10; b = -19; c = +6;
Δ = b2-4ac
Δ = -192-4·10·6
Δ = 121
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{121}=11$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-11}{2*10}=\frac{8}{20} =2/5 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+11}{2*10}=\frac{30}{20} =1+1/2 $
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