Solve (3+2y)^2+2y^2=3

$$( 3 + 2 y ) ^ { 2 } + 2 y ^ { 2 } = 3$$

Solve for y

$$y=-1$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3+2y\right)^{2}.

$$9+12y+4y^{2}+2y^{2}=3$$

Combine 4y^{2} and 2y^{2} to get 6y^{2}.

$$9+12y+6y^{2}=3$$

Subtract 3 from both sides.

$$9+12y+6y^{2}-3=0$$

Subtract 3 from 9 to get 6.

$$6+12y+6y^{2}=0$$

Divide both sides by 6.

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

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

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as y^{2}+ay+by+1. To find a and b, set up a system to be solved.

$$a+b=2 ab=1\times 1=1$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. The only such pair is the system solution.

$$a=1 b=1$$

Rewrite y^{2}+2y+1 as \left(y^{2}+y\right)+\left(y+1\right).

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

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

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

Factor out common term y+1 by using distributive property.

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

Rewrite as a binomial square.

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

To find equation solution, solve y+1=0.

$$y=-1$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3+2y\right)^{2}.

$$9+12y+4y^{2}+2y^{2}=3$$

Combine 4y^{2} and 2y^{2} to get 6y^{2}.

$$9+12y+6y^{2}=3$$

Subtract 3 from both sides.

$$9+12y+6y^{2}-3=0$$

Subtract 3 from 9 to get 6.

$$6+12y+6y^{2}=0$$

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.

$$6y^{2}+12y+6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 6 for a, 12 for b, and 6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-12±\sqrt{12^{2}-4\times 6\times 6}}{2\times 6}$$

Square 12.

$$y=\frac{-12±\sqrt{144-4\times 6\times 6}}{2\times 6}$$

Multiply -4 times 6.

$$y=\frac{-12±\sqrt{144-24\times 6}}{2\times 6}$$

Multiply -24 times 6.

$$y=\frac{-12±\sqrt{144-144}}{2\times 6}$$

Add 144 to -144.

$$y=\frac{-12±\sqrt{0}}{2\times 6}$$

Take the square root of 0.

$$y=-\frac{12}{2\times 6}$$

Multiply 2 times 6.

$$y=-\frac{12}{12}$$

Divide -12 by 12.

$$y=-1$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3+2y\right)^{2}.

$$9+12y+4y^{2}+2y^{2}=3$$

Combine 4y^{2} and 2y^{2} to get 6y^{2}.

$$9+12y+6y^{2}=3$$

Subtract 9 from both sides.

$$12y+6y^{2}=3-9$$

Subtract 9 from 3 to get -6.

$$12y+6y^{2}=-6$$

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.

$$6y^{2}+12y=-6$$

Divide both sides by 6.

$$\frac{6y^{2}+12y}{6}=-\frac{6}{6}$$

Dividing by 6 undoes the multiplication by 6.

$$y^{2}+\frac{12}{6}y=-\frac{6}{6}$$

Divide 12 by 6.

$$y^{2}+2y=-\frac{6}{6}$$

Divide -6 by 6.

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

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.

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

Square 1.

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

Add -1 to 1.

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

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$y+1=0 y+1=0$$

Subtract 1 from both sides of the equation.

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

The equation is now solved. Solutions are the same.

$$y=-1$$