Solve 4(1+2y+y^2)-y^2=4

$$4 ( 1 + 2 y + y ^ { 2 } ) - y ^ { 2 } = 4$$

Solve for y

$$y = -\frac{8}{3} = -2\frac{2}{3} \approx -2.666666667$$

$$y=0$$

Steps Using Factoring

Use the distributive property to multiply 4 by 1+2y+y^{2}.

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

Combine 4y^{2} and -y^{2} to get 3y^{2}.

$$4+8y+3y^{2}=4$$

Subtract 4 from both sides.

$$4+8y+3y^{2}-4=0$$

Subtract 4 from 4 to get 0.

$$8y+3y^{2}=0$$

Factor out y.

$$y\left(8+3y\right)=0$$

To find equation solutions, solve y=0 and 8+3y=0.

$$y=0 y=-\frac{8}{3}$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 4 by 1+2y+y^{2}.

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

Combine 4y^{2} and -y^{2} to get 3y^{2}.

$$4+8y+3y^{2}=4$$

Subtract 4 from both sides.

$$4+8y+3y^{2}-4=0$$

Subtract 4 from 4 to get 0.

$$8y+3y^{2}=0$$

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.

$$3y^{2}+8y=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 8 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-8±\sqrt{8^{2}}}{2\times 3}$$

Take the square root of 8^{2}.

$$y=\frac{-8±8}{2\times 3}$$

Multiply 2 times 3.

$$y=\frac{-8±8}{6}$$

Now solve the equation y=\frac{-8±8}{6} when ± is plus. Add -8 to 8.

$$y=\frac{0}{6}$$

Divide 0 by 6.

$$y=0$$

Now solve the equation y=\frac{-8±8}{6} when ± is minus. Subtract 8 from -8.

$$y=-\frac{16}{6}$$

Reduce the fraction \frac{-16}{6} to lowest terms by extracting and canceling out 2.

$$y=-\frac{8}{3}$$

The equation is now solved.

$$y=0 y=-\frac{8}{3}$$

Steps for Completing the Square

Use the distributive property to multiply 4 by 1+2y+y^{2}.

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

Combine 4y^{2} and -y^{2} to get 3y^{2}.

$$4+8y+3y^{2}=4$$

Subtract 4 from both sides.

$$8y+3y^{2}=4-4$$

Subtract 4 from 4 to get 0.

$$8y+3y^{2}=0$$

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.

$$3y^{2}+8y=0$$

Divide both sides by 3.

$$\frac{3y^{2}+8y}{3}=\frac{0}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$y^{2}+\frac{8}{3}y=\frac{0}{3}$$

Divide 0 by 3.

$$y^{2}+\frac{8}{3}y=0$$

Divide \frac{8}{3}, the coefficient of the x term, by 2 to get \frac{4}{3}. Then add the square of \frac{4}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}+\frac{8}{3}y+\left(\frac{4}{3}\right)^{2}=\left(\frac{4}{3}\right)^{2}$$

Square \frac{4}{3} by squaring both the numerator and the denominator of the fraction.

$$y^{2}+\frac{8}{3}y+\frac{16}{9}=\frac{16}{9}$$

Factor y^{2}+\frac{8}{3}y+\frac{16}{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}.

$$\left(y+\frac{4}{3}\right)^{2}=\frac{16}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y+\frac{4}{3}\right)^{2}}=\sqrt{\frac{16}{9}}$$

Simplify.

$$y+\frac{4}{3}=\frac{4}{3} y+\frac{4}{3}=-\frac{4}{3}$$

Subtract \frac{4}{3} from both sides of the equation.

$$y=0 y=-\frac{8}{3}$$