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3y^2+12y+5=y
We move all terms to the left:
3y^2+12y+5-(y)=0
We add all the numbers together, and all the variables
3y^2+11y+5=0
a = 3; b = 11; c = +5;
Δ = b2-4ac
Δ = 112-4·3·5
Δ = 61
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-\sqrt{61}}{2*3}=\frac{-11-\sqrt{61}}{6} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+\sqrt{61}}{2*3}=\frac{-11+\sqrt{61}}{6} $
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