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12y-4y^2=0
a = -4; b = 12; c = 0;
Δ = b2-4ac
Δ = 122-4·(-4)·0
Δ = 144
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{144}=12$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-12}{2*-4}=\frac{-24}{-8} =+3 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+12}{2*-4}=\frac{0}{-8} =0 $
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