Solve for y (y-3)^2=4y-12

${(y - 3)}^{2} = 4 y - 12$

Step 1

Move all terms containing $y$ to the left side of the equation.

Subtract $4 y$ from both sides of the equation.

${(y - 3)}^{2} - 4 y = - 12$

Simplify each term.

Rewrite ${(y - 3)}^{2}$ as $(y - 3) (y - 3)$ .

$(y - 3) (y - 3) - 4 y = - 12$

Expand $(y - 3) (y - 3)$ using the FOIL Method.

Apply the distributive property.

$y (y - 3) - 3 (y - 3) - 4 y = - 12$

Apply the distributive property.

$y \cdot y + y \cdot - 3 - 3 (y - 3) - 4 y = - 12$

Apply the distributive property.

$y \cdot y + y \cdot - 3 - 3 y - 3 \cdot - 3 - 4 y = - 12$

Simplify and combine like terms.

Simplify each term.

Multiply $y$ by $y$ .

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

Move $- 3$ to the left of $y$ .

$y^{2} - 3 \cdot y - 3 y - 3 \cdot - 3 - 4 y = - 12$

Multiply $- 3$ by $- 3$ .

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

Subtract $3 y$ from $- 3 y$ .

$y^{2} - 6 y + 9 - 4 y = - 12$

Subtract $4 y$ from $- 6 y$ .

$y^{2} - 10 y + 9 = - 12$

Step 2

Add $12$ to both sides of the equation.

$y^{2} - 10 y + 9 + 12 = 0$

Step 3

Add $9$ and $12$ .

$y^{2} - 10 y + 21 = 0$

Step 4

Factor $y^{2} - 10 y + 21$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $21$ and whose sum is $- 10$ .

$- 7, - 3$

Write the factored form using these integers.

$(y - 7) (y - 3) = 0$

Step 5

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$y - 7 = 0$

$y - 3 = 0$

Step 6

Set $y - 7$ equal to $0$ and solve for $y$ .

Set $y - 7$ equal to $0$ .

$y - 7 = 0$

Add $7$ to both sides of the equation.

$y = 7$

Step 7

Set $y - 3$ equal to $0$ and solve for $y$ .

Set $y - 3$ equal to $0$ .

$y - 3 = 0$

Add $3$ to both sides of the equation.

$y = 3$

Step 8

The final solution is all the values that make $(y - 7) (y - 3) = 0$ true.

$y = 7, 3$

Solve for y (y-3)^2=4y-12

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