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3y^2-12y-36=0
a = 3; b = -12; c = -36;
Δ = b2-4ac
Δ = -122-4·3·(-36)
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$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-24}{2*3}=\frac{-12}{6} =-2 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+24}{2*3}=\frac{36}{6} =6 $
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