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6b^2+12b=0
a = 6; b = 12; c = 0;
Δ = b2-4ac
Δ = 122-4·6·0
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{144}=12$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-12}{2*6}=\frac{-24}{12} =-2 $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+12}{2*6}=\frac{0}{12} =0 $
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