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y=2y-y^2
We move all terms to the left:
y-(2y-y^2)=0
We get rid of parentheses
y^2-2y+y=0
We add all the numbers together, and all the variables
y^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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