Solve for x 3x^2-5x=6-2x

$3 x^{2} - 5 x = 6 - 2 x$

Step 1

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

Add $2 x$ to both sides of the equation.

$3 x^{2} - 5 x + 2 x = 6$

Add $- 5 x$ and $2 x$ .

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

Step 2

Subtract $6$ from both sides of the equation.

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

Step 3

Factor the left side of the equation.

Factor $3$ out of $3 x^{2} - 3 x - 6$ .

Factor $3$ out of $3 x^{2}$ .

$3 (x^{2}) - 3 x - 6 = 0$

Factor $3$ out of $- 3 x$ .

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

Factor $3$ out of $- 6$ .

$3 (x^{2}) + 3 (- x) + 3 (- 2) = 0$

Factor $3$ out of $3 (x^{2}) + 3 (- x)$ .

$3 (x^{2} - x) + 3 (- 2) = 0$

Factor $3$ out of $3 (x^{2} - x) + 3 (- 2)$ .

$3 (x^{2} - x - 2) = 0$

Factor.

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

$- 2, 1$

Write the factored form using these integers.

$3 ((x - 2) (x + 1)) = 0$

Remove unnecessary parentheses.

$3 (x - 2) (x + 1) = 0$

Step 4

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

$x - 2 = 0$

$x + 1 = 0$

Step 5

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.

$x = 2$

Step 6

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

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

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 7

The final solution is all the values that make $3 (x - 2) (x + 1) = 0$ true.

$x = 2, - 1$

Solve for x 3x^2-5x=6-2x

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