Solve for x -x^2+6x=x+4

$- x^{2} + 6 x = x + 4$

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

Subtract $x$ from both sides of the equation.

$- x^{2} + 6 x - x = 4$

Subtract $x$ from $6 x$ .

$- x^{2} + 5 x = 4$

Subtract $4$ from both sides of the equation.

$- x^{2} + 5 x - 4 = 0$

Factor the left side of the equation.

Factor $- 1$ out of $- x^{2} + 5 x - 4$ .

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

$- (x^{2}) + 5 x - 4 = 0$

Factor $- 1$ out of $5 x$ .

$- (x^{2}) - (- 5 x) - 4 = 0$

Rewrite $- 4$ as $- 1 (4)$ .

$- (x^{2}) - (- 5 x) - 1 \cdot 4 = 0$

Factor $- 1$ out of $- (x^{2}) - (- 5 x)$ .

$- (x^{2} - 5 x) - 1 \cdot 4 = 0$

Factor $- 1$ out of $- (x^{2} - 5 x) - 1 (4)$ .

$- (x^{2} - 5 x + 4) = 0$

Factor.

Factor $x^{2} - 5 x + 4$ 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 $4$ and whose sum is $- 5$ .

$- 4, - 1$

Write the factored form using these integers.

$- ((x - 4) (x - 1)) = 0$

Remove unnecessary parentheses.

$- (x - 4) (x - 1) = 0$

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

$x - 4 = 0$

$x - 1 = 0$

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.

$x = 4$

Set $x - 1$ equal to $0$ and solve for $x$ .

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

The final solution is all the values that make $- (x - 4) (x - 1) = 0$ true.

$x = 4, 1$

Solve for x -x^2+6x=x+4

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