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-6u^2+11u-3=0
a = -6; b = 11; c = -3;
Δ = b2-4ac
Δ = 112-4·(-6)·(-3)
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{49}=7$$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-7}{2*-6}=\frac{-18}{-12} =1+1/2 $$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+7}{2*-6}=\frac{-4}{-12} =1/3 $
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