Solve (y-1)^2+2y(1-y)=0

$$( y - 1 ) ^ { 2 } + 2 y ( 1 - y ) = 0$$

Solve for y

$$y=1$$

$$y=-1$$

Steps by Finding Square Root

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(y-1\right)^{2}.

$$y^{2}-2y+1+2y\left(1-y\right)=0$$

Use the distributive property to multiply 2y by 1-y.

$$y^{2}-2y+1+2y-2y^{2}=0$$

Combine -2y and 2y to get 0.

$$y^{2}+1-2y^{2}=0$$

Combine y^{2} and -2y^{2} to get -y^{2}.

$$-y^{2}+1=0$$

Subtract 1 from both sides. Anything subtracted from zero gives its negation.

$$-y^{2}=-1$$

Divide both sides by -1.

$$y^{2}=\frac{-1}{-1}$$

Divide -1 by -1 to get 1.

$$y^{2}=1$$

Take the square root of both sides of the equation.

$$y=1 y=-1$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(y-1\right)^{2}.

$$y^{2}-2y+1+2y\left(1-y\right)=0$$

Use the distributive property to multiply 2y by 1-y.

$$y^{2}-2y+1+2y-2y^{2}=0$$

Combine -2y and 2y to get 0.

$$y^{2}+1-2y^{2}=0$$

Combine y^{2} and -2y^{2} to get -y^{2}.

$$-y^{2}+1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 0 for b, and 1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{0±\sqrt{0^{2}-4\left(-1\right)}}{2\left(-1\right)}$$

Square 0.

$$y=\frac{0±\sqrt{-4\left(-1\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$y=\frac{0±\sqrt{4}}{2\left(-1\right)}$$

Take the square root of 4.

$$y=\frac{0±2}{2\left(-1\right)}$$

Multiply 2 times -1.

$$y=\frac{0±2}{-2}$$

Now solve the equation y=\frac{0±2}{-2} when ± is plus. Divide 2 by -2.

$$y=-1$$

Now solve the equation y=\frac{0±2}{-2} when ± is minus. Divide -2 by -2.

$$y=1$$

The equation is now solved.

$$y=-1 y=1$$