Solve left(6-yright)^2+y^2=2(6-y)y+4

$${ \left(6-y \right) }^{ 2 } + { y }^{ 2 } =2(6-y)y+4$$

Solve for y

$$y=2$$

$$y=4$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(6-y\right)^{2}.

$$36-12y+y^{2}+y^{2}=2\left(6-y\right)y+4$$

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

$$36-12y+2y^{2}=2\left(6-y\right)y+4$$

Use the distributive property to multiply 2 by 6-y.

$$36-12y+2y^{2}=\left(12-2y\right)y+4$$

Use the distributive property to multiply 12-2y by y.

$$36-12y+2y^{2}=12y-2y^{2}+4$$

Subtract 12y from both sides.

$$36-12y+2y^{2}-12y=-2y^{2}+4$$

Combine -12y and -12y to get -24y.

$$36-24y+2y^{2}=-2y^{2}+4$$

Add 2y^{2} to both sides.

$$36-24y+2y^{2}+2y^{2}=4$$

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

$$36-24y+4y^{2}=4$$

Subtract 4 from both sides.

$$36-24y+4y^{2}-4=0$$

Subtract 4 from 36 to get 32.

$$32-24y+4y^{2}=0$$

Divide both sides by 4.

$$8-6y+y^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$y^{2}-6y+8=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as y^{2}+ay+by+8. To find a and b, set up a system to be solved.

$$a+b=-6 ab=1\times 8=8$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 8.

$$-1,-8 -2,-4$$

Calculate the sum for each pair.

$$-1-8=-9 -2-4=-6$$

The solution is the pair that gives sum -6.

$$a=-4 b=-2$$

Rewrite y^{2}-6y+8 as \left(y^{2}-4y\right)+\left(-2y+8\right).

$$\left(y^{2}-4y\right)+\left(-2y+8\right)$$

Factor out y in the first and -2 in the second group.

$$y\left(y-4\right)-2\left(y-4\right)$$

Factor out common term y-4 by using distributive property.

$$\left(y-4\right)\left(y-2\right)$$

To find equation solutions, solve y-4=0 and y-2=0.

$$y=4 y=2$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(6-y\right)^{2}.

$$36-12y+y^{2}+y^{2}=2\left(6-y\right)y+4$$

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

$$36-12y+2y^{2}=2\left(6-y\right)y+4$$

Use the distributive property to multiply 2 by 6-y.

$$36-12y+2y^{2}=\left(12-2y\right)y+4$$

Use the distributive property to multiply 12-2y by y.

$$36-12y+2y^{2}=12y-2y^{2}+4$$

Subtract 12y from both sides.

$$36-12y+2y^{2}-12y=-2y^{2}+4$$

Combine -12y and -12y to get -24y.

$$36-24y+2y^{2}=-2y^{2}+4$$

Add 2y^{2} to both sides.

$$36-24y+2y^{2}+2y^{2}=4$$

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

$$36-24y+4y^{2}=4$$

Subtract 4 from both sides.

$$36-24y+4y^{2}-4=0$$

Subtract 4 from 36 to get 32.

$$32-24y+4y^{2}=0$$

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.

$$4y^{2}-24y+32=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -24 for b, and 32 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-24\right)±\sqrt{\left(-24\right)^{2}-4\times 4\times 32}}{2\times 4}$$

Square -24.

$$y=\frac{-\left(-24\right)±\sqrt{576-4\times 4\times 32}}{2\times 4}$$

Multiply -4 times 4.

$$y=\frac{-\left(-24\right)±\sqrt{576-16\times 32}}{2\times 4}$$

Multiply -16 times 32.

$$y=\frac{-\left(-24\right)±\sqrt{576-512}}{2\times 4}$$

Add 576 to -512.

$$y=\frac{-\left(-24\right)±\sqrt{64}}{2\times 4}$$

Take the square root of 64.

$$y=\frac{-\left(-24\right)±8}{2\times 4}$$

The opposite of -24 is 24.

$$y=\frac{24±8}{2\times 4}$$

Multiply 2 times 4.

$$y=\frac{24±8}{8}$$

Now solve the equation y=\frac{24±8}{8} when ± is plus. Add 24 to 8.

$$y=\frac{32}{8}$$

Divide 32 by 8.

$$y=4$$

Now solve the equation y=\frac{24±8}{8} when ± is minus. Subtract 8 from 24.

$$y=\frac{16}{8}$$

Divide 16 by 8.

$$y=2$$

The equation is now solved.

$$y=4 y=2$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(6-y\right)^{2}.

$$36-12y+y^{2}+y^{2}=2\left(6-y\right)y+4$$

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

$$36-12y+2y^{2}=2\left(6-y\right)y+4$$

Use the distributive property to multiply 2 by 6-y.

$$36-12y+2y^{2}=\left(12-2y\right)y+4$$

Use the distributive property to multiply 12-2y by y.

$$36-12y+2y^{2}=12y-2y^{2}+4$$

Subtract 12y from both sides.

$$36-12y+2y^{2}-12y=-2y^{2}+4$$

Combine -12y and -12y to get -24y.

$$36-24y+2y^{2}=-2y^{2}+4$$

Add 2y^{2} to both sides.

$$36-24y+2y^{2}+2y^{2}=4$$

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

$$36-24y+4y^{2}=4$$

Subtract 36 from both sides.

$$-24y+4y^{2}=4-36$$

Subtract 36 from 4 to get -32.

$$-24y+4y^{2}=-32$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$4y^{2}-24y=-32$$

Divide both sides by 4.

$$\frac{4y^{2}-24y}{4}=-\frac{32}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$y^{2}+\left(-\frac{24}{4}\right)y=-\frac{32}{4}$$

Divide -24 by 4.

$$y^{2}-6y=-\frac{32}{4}$$

Divide -32 by 4.

$$y^{2}-6y=-8$$

Divide -6, the coefficient of the x term, by 2 to get -3. Then add the square of -3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}-6y+\left(-3\right)^{2}=-8+\left(-3\right)^{2}$$

Square -3.

$$y^{2}-6y+9=-8+9$$

Add -8 to 9.

$$y^{2}-6y+9=1$$

Factor y^{2}-6y+9. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$y-3=1 y-3=-1$$

Add 3 to both sides of the equation.

$$y=4 y=2$$