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=Y^2+Y
We move all terms to the left:
-(Y^2+Y)=0
We get rid of parentheses
-Y^2-Y=0
We add all the numbers together, and all the variables
-1Y^2-1Y=0
a = -1; b = -1; c = 0;
Δ = b2-4ac
Δ = -12-4·(-1)·0
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$$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-1}{2*-1}=\frac{0}{-2} =0 $$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+1}{2*-1}=\frac{2}{-2} =-1 $
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