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=2Y^2-3Y-2
We move all terms to the left:
-(2Y^2-3Y-2)=0
We get rid of parentheses
-2Y^2+3Y+2=0
a = -2; b = 3; c = +2;
Δ = b2-4ac
Δ = 32-4·(-2)·2
Δ = 25
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{25}=5$$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-5}{2*-2}=\frac{-8}{-4} =+2 $$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+5}{2*-2}=\frac{2}{-4} =-1/2 $
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