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11d^2-33d=0
a = 11; b = -33; c = 0;
Δ = b2-4ac
Δ = -332-4·11·0
Δ = 1089
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{1089}=33$$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-33)-33}{2*11}=\frac{0}{22} =0 $$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-33)+33}{2*11}=\frac{66}{22} =3 $
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