Solve left(4-2right)^2+left(y+1right)^2=16

$${ \left(4-2 \right) }^{ 2 } + { \left(y+1 \right) }^{ 2 } =16$$

Solve for y

$$y=2\sqrt{3}-1\approx 2.464101615$$

$$y=-2\sqrt{3}-1\approx -4.464101615$$

Steps Using the Quadratic Formula

Subtract 2 from 4 to get 2.

$$2^{2}+\left(y+1\right)^{2}=16$$

Calculate 2 to the power of 2 and get 4.

$$4+\left(y+1\right)^{2}=16$$

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

$$4+y^{2}+2y+1=16$$

Add 4 and 1 to get 5.

$$5+y^{2}+2y=16$$

Subtract 16 from both sides.

$$5+y^{2}+2y-16=0$$

Subtract 16 from 5 to get -11.

$$-11+y^{2}+2y=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$y^{2}+2y-11=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 2 for b, and -11 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-2±\sqrt{2^{2}-4\left(-11\right)}}{2}$$

Square 2.

$$y=\frac{-2±\sqrt{4-4\left(-11\right)}}{2}$$

Multiply -4 times -11.

$$y=\frac{-2±\sqrt{4+44}}{2}$$

Add 4 to 44.

$$y=\frac{-2±\sqrt{48}}{2}$$

Take the square root of 48.

$$y=\frac{-2±4\sqrt{3}}{2}$$

Now solve the equation y=\frac{-2±4\sqrt{3}}{2} when ± is plus. Add -2 to 4\sqrt{3}.

$$y=\frac{4\sqrt{3}-2}{2}$$

Divide -2+4\sqrt{3} by 2.

$$y=2\sqrt{3}-1$$

Now solve the equation y=\frac{-2±4\sqrt{3}}{2} when ± is minus. Subtract 4\sqrt{3} from -2.

$$y=\frac{-4\sqrt{3}-2}{2}$$

Divide -2-4\sqrt{3} by 2.

$$y=-2\sqrt{3}-1$$

The equation is now solved.

$$y=2\sqrt{3}-1 y=-2\sqrt{3}-1$$

Steps for Completing the Square

Subtract 2 from 4 to get 2.

$$2^{2}+\left(y+1\right)^{2}=16$$

Calculate 2 to the power of 2 and get 4.

$$4+\left(y+1\right)^{2}=16$$

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

$$4+y^{2}+2y+1=16$$

Add 4 and 1 to get 5.

$$5+y^{2}+2y=16$$

Subtract 5 from both sides.

$$y^{2}+2y=16-5$$

Subtract 5 from 16 to get 11.

$$y^{2}+2y=11$$

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

$$y^{2}+2y+1^{2}=11+1^{2}$$

Square 1.

$$y^{2}+2y+1=11+1$$

Add 11 to 1.

$$y^{2}+2y+1=12$$

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

$$\left(y+1\right)^{2}=12$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y+1\right)^{2}}=\sqrt{12}$$

Simplify.

$$y+1=2\sqrt{3} y+1=-2\sqrt{3}$$

Subtract 1 from both sides of the equation.

$$y=2\sqrt{3}-1 y=-2\sqrt{3}-1$$