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12y^2+24y=0
a = 12; b = 24; c = 0;
Δ = b2-4ac
Δ = 242-4·12·0
Δ = 576
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{576}=24$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-24}{2*12}=\frac{-48}{24} =-2 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+24}{2*12}=\frac{0}{24} =0 $
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