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3-12y^2=0
a = -12; b = 0; c = +3;
Δ = b2-4ac
Δ = 02-4·(-12)·3
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$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12}{2*-12}=\frac{-12}{-24} =1/2 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12}{2*-12}=\frac{12}{-24} =-1/2 $
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