Solve for x 12-x^2+x=0

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

Step 1

Factor the left side of the equation.

Factor $- 1$ out of $12 - x^{2} + x$ .

Move $12$ .

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

Factor $- 1$ out of $- x^{2}$ .

$- (x^{2}) + x + 12 = 0$

Factor $- 1$ out of $x$ .

$- (x^{2}) - 1 (- x) + 12 = 0$

Rewrite $12$ as $- 1 (- 12)$ .

$- (x^{2}) - 1 (- x) - 1 \cdot - 12 = 0$

Factor $- 1$ out of $- (x^{2}) - 1 (- x)$ .

$- (x^{2} - x) - 1 \cdot - 12 = 0$

Factor $- 1$ out of $- (x^{2} - x) - 1 (- 12)$ .

$- (x^{2} - x - 12) = 0$

Factor.

Factor $x^{2} - x - 12$ 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 $- 12$ and whose sum is $- 1$ .

$- 4, 3$

Write the factored form using these integers.

$- ((x - 4) (x + 3)) = 0$

Remove unnecessary parentheses.

$- (x - 4) (x + 3) = 0$

Step 2

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

$x - 4 = 0$

$x + 3 = 0$

Step 3

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.

$x = 4$

Step 4

Set $x + 3$ equal to $0$ and solve for $x$ .

Set $x + 3$ equal to $0$ .

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 5

The final solution is all the values that make $- (x - 4) (x + 3) = 0$ true.

$x = 4, - 3$

Solve for x 12-x^2+x=0

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