Solve for x x^2+6x-4=3x+6

$x^{2} + 6 x - 4 = 3 x + 6$

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

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

$x^{2} + 6 x - 4 - 3 x = 6$

Subtract $3 x$ from $6 x$ .

$x^{2} + 3 x - 4 = 6$

Subtract $6$ from both sides of the equation.

$x^{2} + 3 x - 4 - 6 = 0$

Subtract $6$ from $- 4$ .

$x^{2} + 3 x - 10 = 0$

Factor $x^{2} + 3 x - 10$ 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 $- 10$ and whose sum is $3$ .

$- 2, 5$

Write the factored form using these integers.

$(x - 2) (x + 5) = 0$

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

$x - 2 = 0$

$x + 5 = 0$

Set $x - 2$ equal to $0$ and solve for $x$ .

Set $x - 2$ equal to $0$ .

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

Set $x + 5$ equal to $0$ and solve for $x$ .

Set $x + 5$ equal to $0$ .

$x + 5 = 0$

Subtract $5$ from both sides of the equation.

$x = - 5$

The final solution is all the values that make $(x - 2) (x + 5) = 0$ true.

$x = 2, - 5$

Solve for x x^2+6x-4=3x+6

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