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

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

Solve for y

$$y=\frac{1}{3}\approx 0.333333333$$

$$y=5$$

Steps Using Factoring By Grouping

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

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

Subtract y^{2} from both sides.

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

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

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

Subtract 4y from both sides.

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

Combine -12y and -4y to get -16y.

$$3y^{2}-16y+9=4$$

Subtract 4 from both sides.

$$3y^{2}-16y+9-4=0$$

Subtract 4 from 9 to get 5.

$$3y^{2}-16y+5=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3y^{2}+ay+by+5. To find a and b, set up a system to be solved.

$$a+b=-16 ab=3\times 5=15$$

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 15.

$$-1,-15 -3,-5$$

Calculate the sum for each pair.

$$-1-15=-16 -3-5=-8$$

The solution is the pair that gives sum -16.

$$a=-15 b=-1$$

Rewrite 3y^{2}-16y+5 as \left(3y^{2}-15y\right)+\left(-y+5\right).

$$\left(3y^{2}-15y\right)+\left(-y+5\right)$$

Factor out 3y in the first and -1 in the second group.

$$3y\left(y-5\right)-\left(y-5\right)$$

Factor out common term y-5 by using distributive property.

$$\left(y-5\right)\left(3y-1\right)$$

To find equation solutions, solve y-5=0 and 3y-1=0.

$$y=5 y=\frac{1}{3}$$

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}.

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

Subtract y^{2} from both sides.

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

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

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

Subtract 4y from both sides.

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

Combine -12y and -4y to get -16y.

$$3y^{2}-16y+9=4$$

Subtract 4 from both sides.

$$3y^{2}-16y+9-4=0$$

Subtract 4 from 9 to get 5.

$$3y^{2}-16y+5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -16 for b, and 5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-16\right)±\sqrt{\left(-16\right)^{2}-4\times 3\times 5}}{2\times 3}$$

Square -16.

$$y=\frac{-\left(-16\right)±\sqrt{256-4\times 3\times 5}}{2\times 3}$$

Multiply -4 times 3.

$$y=\frac{-\left(-16\right)±\sqrt{256-12\times 5}}{2\times 3}$$

Multiply -12 times 5.

$$y=\frac{-\left(-16\right)±\sqrt{256-60}}{2\times 3}$$

Add 256 to -60.

$$y=\frac{-\left(-16\right)±\sqrt{196}}{2\times 3}$$

Take the square root of 196.

$$y=\frac{-\left(-16\right)±14}{2\times 3}$$

The opposite of -16 is 16.

$$y=\frac{16±14}{2\times 3}$$

Multiply 2 times 3.

$$y=\frac{16±14}{6}$$

Now solve the equation y=\frac{16±14}{6} when ± is plus. Add 16 to 14.

$$y=\frac{30}{6}$$

Divide 30 by 6.

$$y=5$$

Now solve the equation y=\frac{16±14}{6} when ± is minus. Subtract 14 from 16.

$$y=\frac{2}{6}$$

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

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

The equation is now solved.

$$y=5 y=\frac{1}{3}$$

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}.

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

Subtract y^{2} from both sides.

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

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

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

Subtract 4y from both sides.

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

Combine -12y and -4y to get -16y.

$$3y^{2}-16y+9=4$$

Subtract 9 from both sides.

$$3y^{2}-16y=4-9$$

Subtract 9 from 4 to get -5.

$$3y^{2}-16y=-5$$

Divide both sides by 3.

$$\frac{3y^{2}-16y}{3}=-\frac{5}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$y^{2}-\frac{16}{3}y=-\frac{5}{3}$$

Divide -\frac{16}{3}, the coefficient of the x term, by 2 to get -\frac{8}{3}. Then add the square of -\frac{8}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}-\frac{16}{3}y+\left(-\frac{8}{3}\right)^{2}=-\frac{5}{3}+\left(-\frac{8}{3}\right)^{2}$$

Square -\frac{8}{3} by squaring both the numerator and the denominator of the fraction.

$$y^{2}-\frac{16}{3}y+\frac{64}{9}=-\frac{5}{3}+\frac{64}{9}$$

Add -\frac{5}{3} to \frac{64}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$y^{2}-\frac{16}{3}y+\frac{64}{9}=\frac{49}{9}$$

Factor y^{2}-\frac{16}{3}y+\frac{64}{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}.

$$\left(y-\frac{8}{3}\right)^{2}=\frac{49}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y-\frac{8}{3}\right)^{2}}=\sqrt{\frac{49}{9}}$$

Simplify.

$$y-\frac{8}{3}=\frac{7}{3} y-\frac{8}{3}=-\frac{7}{3}$$

Add \frac{8}{3} to both sides of the equation.

$$y=5 y=\frac{1}{3}$$