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

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

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

Subtract $3 x$ from both sides of the equation.

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

Simplify each term.

Rewrite ${(x - 4)}^{2}$ as $(x - 4) (x - 4)$ .

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

Expand $(x - 4) (x - 4)$ using the FOIL Method.

Apply the distributive property.

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

Apply the distributive property.

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

Apply the distributive property.

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

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

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

Move $- 4$ to the left of $x$ .

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

Multiply $- 4$ by $- 4$ .

$x^{2} - 4 x - 4 x + 16 - 3 x = - 12$

Subtract $4 x$ from $- 4 x$ .

$x^{2} - 8 x + 16 - 3 x = - 12$

Subtract $3 x$ from $- 8 x$ .

$x^{2} - 11 x + 16 = - 12$

Add $12$ to both sides of the equation.

$x^{2} - 11 x + 16 + 12 = 0$

Add $16$ and $12$ .

$x^{2} - 11 x + 28 = 0$

Factor $x^{2} - 11 x + 28$ 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 $28$ and whose sum is $- 11$ .

$- 7, - 4$

Write the factored form using these integers.

$(x - 7) (x - 4) = 0$

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

$x - 7 = 0$

$x - 4 = 0$

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

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

$x - 7 = 0$

Add $7$ to both sides of the equation.

$x = 7$

Set $x - 4$ equal to $0$ and solve for $x$ .

Set $x - 4$ equal to $0$ .

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

The final solution is all the values that make $(x - 7) (x - 4) = 0$ true.

$x = 7, 4$

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

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