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3d^2-11d=0
a = 3; b = -11; c = 0;
Δ = b2-4ac
Δ = -112-4·3·0
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{121}=11$$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-11}{2*3}=\frac{0}{6} =0 $$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+11}{2*3}=\frac{22}{6} =3+2/3 $
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