Solve (2y+3)^2+y^2=4

$$( 2 y + 3 ) ^ { 2 } + y ^ { 2 } = 4$$

Solve for y

$$y=\frac{\sqrt{11}-6}{5}\approx -0.536675042$$

$$y=\frac{-\sqrt{11}-6}{5}\approx -1.863324958$$

Steps Using the Quadratic Formula

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

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

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

$$5y^{2}+12y+9=4$$

Subtract 4 from both sides.

$$5y^{2}+12y+9-4=0$$

Subtract 4 from 9 to get 5.

$$5y^{2}+12y+5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, 12 for b, and 5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-12±\sqrt{12^{2}-4\times 5\times 5}}{2\times 5}$$

Square 12.

$$y=\frac{-12±\sqrt{144-4\times 5\times 5}}{2\times 5}$$

Multiply -4 times 5.

$$y=\frac{-12±\sqrt{144-20\times 5}}{2\times 5}$$

Multiply -20 times 5.

$$y=\frac{-12±\sqrt{144-100}}{2\times 5}$$

Add 144 to -100.

$$y=\frac{-12±\sqrt{44}}{2\times 5}$$

Take the square root of 44.

$$y=\frac{-12±2\sqrt{11}}{2\times 5}$$

Multiply 2 times 5.

$$y=\frac{-12±2\sqrt{11}}{10}$$

Now solve the equation y=\frac{-12±2\sqrt{11}}{10} when ± is plus. Add -12 to 2\sqrt{11}.

$$y=\frac{2\sqrt{11}-12}{10}$$

Divide -12+2\sqrt{11} by 10.

$$y=\frac{\sqrt{11}-6}{5}$$

Now solve the equation y=\frac{-12±2\sqrt{11}}{10} when ± is minus. Subtract 2\sqrt{11} from -12.

$$y=\frac{-2\sqrt{11}-12}{10}$$

Divide -12-2\sqrt{11} by 10.

$$y=\frac{-\sqrt{11}-6}{5}$$

The equation is now solved.

$$y=\frac{\sqrt{11}-6}{5} y=\frac{-\sqrt{11}-6}{5}$$

Steps for Completing the Square

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

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

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

$$5y^{2}+12y+9=4$$

Subtract 9 from both sides.

$$5y^{2}+12y=4-9$$

Subtract 9 from 4 to get -5.

$$5y^{2}+12y=-5$$

Divide both sides by 5.

$$\frac{5y^{2}+12y}{5}=-\frac{5}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$y^{2}+\frac{12}{5}y=-\frac{5}{5}$$

Divide -5 by 5.

$$y^{2}+\frac{12}{5}y=-1$$

Divide \frac{12}{5}, the coefficient of the x term, by 2 to get \frac{6}{5}. Then add the square of \frac{6}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}+\frac{12}{5}y+\left(\frac{6}{5}\right)^{2}=-1+\left(\frac{6}{5}\right)^{2}$$

Square \frac{6}{5} by squaring both the numerator and the denominator of the fraction.

$$y^{2}+\frac{12}{5}y+\frac{36}{25}=-1+\frac{36}{25}$$

Add -1 to \frac{36}{25}.

$$y^{2}+\frac{12}{5}y+\frac{36}{25}=\frac{11}{25}$$

Factor y^{2}+\frac{12}{5}y+\frac{36}{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}.

$$\left(y+\frac{6}{5}\right)^{2}=\frac{11}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y+\frac{6}{5}\right)^{2}}=\sqrt{\frac{11}{25}}$$

Simplify.

$$y+\frac{6}{5}=\frac{\sqrt{11}}{5} y+\frac{6}{5}=-\frac{\sqrt{11}}{5}$$

Subtract \frac{6}{5} from both sides of the equation.

$$y=\frac{\sqrt{11}-6}{5} y=\frac{-\sqrt{11}-6}{5}$$