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6x^2-31x+40=0
a = 6; b = -31; c = +40;
Δ = b2-4ac
Δ = -312-4·6·40
Δ = 1
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{1}=1$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-31)-1}{2*6}=\frac{30}{12} =2+1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-31)+1}{2*6}=\frac{32}{12} =2+2/3 $
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