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-3u(u-2)=0
We multiply parentheses
-3u^2+6u=0
a = -3; b = 6; c = 0;
Δ = b2-4ac
Δ = 62-4·(-3)·0
Δ = 36
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{36}=6$$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-6}{2*-3}=\frac{-12}{-6} =+2 $$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6}{2*-3}=\frac{0}{-6} =0 $
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