Solve left(-4-yright)^2+y^2+6(-4-y)-4y+4=0

$${ \left(-4-y \right) }^{ 2 } + { y }^{ 2 } +6(-4-y)-4y+4=0$$

Solve for y

$$y=-1$$

$$y=2$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(-4-y\right)^{2}.

$$16+8y+y^{2}+y^{2}+6\left(-4-y\right)-4y+4=0$$

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

$$16+8y+2y^{2}+6\left(-4-y\right)-4y+4=0$$

Use the distributive property to multiply 6 by -4-y.

$$16+8y+2y^{2}-24-6y-4y+4=0$$

Subtract 24 from 16 to get -8.

$$-8+8y+2y^{2}-6y-4y+4=0$$

Combine 8y and -6y to get 2y.

$$-8+2y+2y^{2}-4y+4=0$$

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

$$-8-2y+2y^{2}+4=0$$

Add -8 and 4 to get -4.

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

Divide both sides by 2.

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

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

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

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

$$a+b=-1 ab=1\left(-2\right)=-2$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. The only such pair is the system solution.

$$a=-2 b=1$$

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

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

Factor out y in y^{2}-2y.

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

Factor out common term y-2 by using distributive property.

$$\left(y-2\right)\left(y+1\right)$$

To find equation solutions, solve y-2=0 and y+1=0.

$$y=2 y=-1$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(-4-y\right)^{2}.

$$16+8y+y^{2}+y^{2}+6\left(-4-y\right)-4y+4=0$$

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

$$16+8y+2y^{2}+6\left(-4-y\right)-4y+4=0$$

Use the distributive property to multiply 6 by -4-y.

$$16+8y+2y^{2}-24-6y-4y+4=0$$

Subtract 24 from 16 to get -8.

$$-8+8y+2y^{2}-6y-4y+4=0$$

Combine 8y and -6y to get 2y.

$$-8+2y+2y^{2}-4y+4=0$$

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

$$-8-2y+2y^{2}+4=0$$

Add -8 and 4 to get -4.

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

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.

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

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

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

Square -2.

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

Multiply -4 times 2.

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

Multiply -8 times -4.

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

Add 4 to 32.

$$y=\frac{-\left(-2\right)±\sqrt{36}}{2\times 2}$$

Take the square root of 36.

$$y=\frac{-\left(-2\right)±6}{2\times 2}$$

The opposite of -2 is 2.

$$y=\frac{2±6}{2\times 2}$$

Multiply 2 times 2.

$$y=\frac{2±6}{4}$$

Now solve the equation y=\frac{2±6}{4} when ± is plus. Add 2 to 6.

$$y=\frac{8}{4}$$

Divide 8 by 4.

$$y=2$$

Now solve the equation y=\frac{2±6}{4} when ± is minus. Subtract 6 from 2.

$$y=-\frac{4}{4}$$

Divide -4 by 4.

$$y=-1$$

The equation is now solved.

$$y=2 y=-1$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(-4-y\right)^{2}.

$$16+8y+y^{2}+y^{2}+6\left(-4-y\right)-4y+4=0$$

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

$$16+8y+2y^{2}+6\left(-4-y\right)-4y+4=0$$

Use the distributive property to multiply 6 by -4-y.

$$16+8y+2y^{2}-24-6y-4y+4=0$$

Subtract 24 from 16 to get -8.

$$-8+8y+2y^{2}-6y-4y+4=0$$

Combine 8y and -6y to get 2y.

$$-8+2y+2y^{2}-4y+4=0$$

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

$$-8-2y+2y^{2}+4=0$$

Add -8 and 4 to get -4.

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

Add 4 to both sides. Anything plus zero gives itself.

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

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.

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

Divide both sides by 2.

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

Dividing by 2 undoes the multiplication by 2.

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

Divide -2 by 2.

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

Divide 4 by 2.

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

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

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

Square -\frac{1}{2} by squaring both the numerator and the denominator of the fraction.

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

Add 2 to \frac{1}{4}.

$$y^{2}-y+\frac{1}{4}=\frac{9}{4}$$

Factor y^{2}-y+\frac{1}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(y-\frac{1}{2}\right)^{2}=\frac{9}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y-\frac{1}{2}\right)^{2}}=\sqrt{\frac{9}{4}}$$

Simplify.

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

Add \frac{1}{2} to both sides of the equation.

$$y=2 y=-1$$