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=2Y+3Y^2
We move all terms to the left:
-(2Y+3Y^2)=0
We get rid of parentheses
-3Y^2-2Y=0
a = -3; b = -2; c = 0;
Δ = b2-4ac
Δ = -22-4·(-3)·0
Δ = 4
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{4}=2$$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2}{2*-3}=\frac{0}{-6} =0 $$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2}{2*-3}=\frac{4}{-6} =-2/3 $
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