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6r^2-11r=0
a = 6; b = -11; c = 0;
Δ = b2-4ac
Δ = -112-4·6·0
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{121}=11$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-11}{2*6}=\frac{0}{12} =0 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+11}{2*6}=\frac{22}{12} =1+5/6 $
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