Solve left(-2y+3-2right)^2+left(y+1right)^2=4

$${ \left(-2y+3-2 \right) }^{ 2 } + { \left(y+1 \right) }^{ 2 } =4$$

Solve for y

$$y=\frac{\sqrt{11}+1}{5}\approx 0.863324958$$

$$y=\frac{1-\sqrt{11}}{5}\approx -0.463324958$$

Steps Using the Quadratic Formula

Subtract 2 from 3 to get 1.

$$\left(-2y+1\right)^{2}+\left(y+1\right)^{2}=4$$

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

$$4y^{2}-4y+1+\left(y+1\right)^{2}=4$$

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

$$4y^{2}-4y+1+y^{2}+2y+1=4$$

Combine 4y^{2} and y^{2} to get 5y^{2}.

$$5y^{2}-4y+1+2y+1=4$$

Combine -4y and 2y to get -2y.

$$5y^{2}-2y+1+1=4$$

Add 1 and 1 to get 2.

$$5y^{2}-2y+2=4$$

Subtract 4 from both sides.

$$5y^{2}-2y+2-4=0$$

Subtract 4 from 2 to get -2.

$$5y^{2}-2y-2=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -2 for b, and -2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\times 5\left(-2\right)}}{2\times 5}$$

Square -2.

$$y=\frac{-\left(-2\right)±\sqrt{4-4\times 5\left(-2\right)}}{2\times 5}$$

Multiply -4 times 5.

$$y=\frac{-\left(-2\right)±\sqrt{4-20\left(-2\right)}}{2\times 5}$$

Multiply -20 times -2.

$$y=\frac{-\left(-2\right)±\sqrt{4+40}}{2\times 5}$$

Add 4 to 40.

$$y=\frac{-\left(-2\right)±\sqrt{44}}{2\times 5}$$

Take the square root of 44.

$$y=\frac{-\left(-2\right)±2\sqrt{11}}{2\times 5}$$

The opposite of -2 is 2.

$$y=\frac{2±2\sqrt{11}}{2\times 5}$$

Multiply 2 times 5.

$$y=\frac{2±2\sqrt{11}}{10}$$

Now solve the equation y=\frac{2±2\sqrt{11}}{10} when ± is plus. Add 2 to 2\sqrt{11}.

$$y=\frac{2\sqrt{11}+2}{10}$$

Divide 2+2\sqrt{11} by 10.

$$y=\frac{\sqrt{11}+1}{5}$$

Now solve the equation y=\frac{2±2\sqrt{11}}{10} when ± is minus. Subtract 2\sqrt{11} from 2.

$$y=\frac{2-2\sqrt{11}}{10}$$

Divide 2-2\sqrt{11} by 10.

$$y=\frac{1-\sqrt{11}}{5}$$

The equation is now solved.

$$y=\frac{\sqrt{11}+1}{5} y=\frac{1-\sqrt{11}}{5}$$

Steps for Completing the Square

Subtract 2 from 3 to get 1.

$$\left(-2y+1\right)^{2}+\left(y+1\right)^{2}=4$$

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

$$4y^{2}-4y+1+\left(y+1\right)^{2}=4$$

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

$$4y^{2}-4y+1+y^{2}+2y+1=4$$

Combine 4y^{2} and y^{2} to get 5y^{2}.

$$5y^{2}-4y+1+2y+1=4$$

Combine -4y and 2y to get -2y.

$$5y^{2}-2y+1+1=4$$

Add 1 and 1 to get 2.

$$5y^{2}-2y+2=4$$

Subtract 2 from both sides.

$$5y^{2}-2y=4-2$$

Subtract 2 from 4 to get 2.

$$5y^{2}-2y=2$$

Divide both sides by 5.

$$\frac{5y^{2}-2y}{5}=\frac{2}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$y^{2}-\frac{2}{5}y=\frac{2}{5}$$

Divide -\frac{2}{5}, the coefficient of the x term, by 2 to get -\frac{1}{5}. Then add the square of -\frac{1}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}-\frac{2}{5}y+\left(-\frac{1}{5}\right)^{2}=\frac{2}{5}+\left(-\frac{1}{5}\right)^{2}$$

Square -\frac{1}{5} by squaring both the numerator and the denominator of the fraction.

$$y^{2}-\frac{2}{5}y+\frac{1}{25}=\frac{2}{5}+\frac{1}{25}$$

Add \frac{2}{5} to \frac{1}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$y^{2}-\frac{2}{5}y+\frac{1}{25}=\frac{11}{25}$$

Factor y^{2}-\frac{2}{5}y+\frac{1}{25}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(y-\frac{1}{5}\right)^{2}=\frac{11}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y-\frac{1}{5}\right)^{2}}=\sqrt{\frac{11}{25}}$$

Simplify.

$$y-\frac{1}{5}=\frac{\sqrt{11}}{5} y-\frac{1}{5}=-\frac{\sqrt{11}}{5}$$

Add \frac{1}{5} to both sides of the equation.

$$y=\frac{\sqrt{11}+1}{5} y=\frac{1-\sqrt{11}}{5}$$