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6x^2-13x+5=0
a = 6; b = -13; c = +5;
Δ = b2-4ac
Δ = -132-4·6·5
Δ = 49
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{49}=7$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-7}{2*6}=\frac{6}{12} =1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+7}{2*6}=\frac{20}{12} =1+2/3 $
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