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6b^2+10b=0
a = 6; b = 10; c = 0;
Δ = b2-4ac
Δ = 102-4·6·0
Δ = 100
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{100}=10$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-10}{2*6}=\frac{-20}{12} =-1+2/3 $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+10}{2*6}=\frac{0}{12} =0 $
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