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3y^2-96+y=360
We move all terms to the left:
3y^2-96+y-(360)=0
We add all the numbers together, and all the variables
3y^2+y-456=0
a = 3; b = 1; c = -456;
Δ = b2-4ac
Δ = 12-4·3·(-456)
Δ = 5473
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{-(1)-\sqrt{5473}}{2*3}=\frac{-1-\sqrt{5473}}{6} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{5473}}{2*3}=\frac{-1+\sqrt{5473}}{6} $
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