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3y-2y^2=-6
We move all terms to the left:
3y-2y^2-(-6)=0
We add all the numbers together, and all the variables
-2y^2+3y+6=0
a = -2; b = 3; c = +6;
Δ = b2-4ac
Δ = 32-4·(-2)·6
Δ = 57
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{-(3)-\sqrt{57}}{2*-2}=\frac{-3-\sqrt{57}}{-4} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{57}}{2*-2}=\frac{-3+\sqrt{57}}{-4} $
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