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3y^2-24y+36=0
a = 3; b = -24; c = +36;
Δ = b2-4ac
Δ = -242-4·3·36
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$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-12}{2*3}=\frac{12}{6} =2 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+12}{2*3}=\frac{36}{6} =6 $
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