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

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

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

Subtract $3 x^{2}$ from both sides of the equation.

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

Subtract $3 x^{2}$ from $2 x^{2}$ .

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

Subtract $3$ from both sides of the equation.

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

Factor the left side of the equation.

Factor $- 1$ out of $- x^{2} + 4 x - 3$ .

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

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

Factor $- 1$ out of $4 x$ .

$- (x^{2}) - (- 4 x) - 3 = 0$

Rewrite $- 3$ as $- 1 (3)$ .

$- (x^{2}) - (- 4 x) - 1 \cdot 3 = 0$

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

$- (x^{2} - 4 x) - 1 \cdot 3 = 0$

Factor $- 1$ out of $- (x^{2} - 4 x) - 1 (3)$ .

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

Factor.

Factor $x^{2} - 4 x + 3$ 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 $3$ and whose sum is $- 4$ .

$- 3, - 1$

Write the factored form using these integers.

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

Remove unnecessary parentheses.

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

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

$x - 3 = 0$

$x - 1 = 0$

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

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

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

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 $- (x - 3) (x - 1) = 0$ true.

$x = 3, 1$

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

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