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6u^2+20u=0
a = 6; b = 20; c = 0;
Δ = b2-4ac
Δ = 202-4·6·0
Δ = 400
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{400}=20$$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-20}{2*6}=\frac{-40}{12} =-3+1/3 $$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+20}{2*6}=\frac{0}{12} =0 $
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