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