Solve for x 4x^2-28=3x^2-3x

$4 x^{2} - 28 = 3 x^{2} - 3 x$

Step 1

Since $x$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

$3 x^{2} - 3 x = 4 x^{2} - 28$

Step 2

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

Subtract $4 x^{2}$ from both sides of the equation.

$3 x^{2} - 3 x - 4 x^{2} = - 28$

Subtract $4 x^{2}$ from $3 x^{2}$ .

$- x^{2} - 3 x = - 28$

Step 3

Add $28$ to both sides of the equation.

$- x^{2} - 3 x + 28 = 0$

Step 4

Factor the left side of the equation.

Factor $- 1$ out of $- x^{2} - 3 x + 28$ .

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

$- (x^{2}) - 3 x + 28 = 0$

Factor $- 1$ out of $- 3 x$ .

$- (x^{2}) - (3 x) + 28 = 0$

Rewrite $28$ as $- 1 (- 28)$ .

$- (x^{2}) - (3 x) - 1 \cdot - 28 = 0$

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

$- (x^{2} + 3 x) - 1 \cdot - 28 = 0$

Factor $- 1$ out of $- (x^{2} + 3 x) - 1 (- 28)$ .

$- (x^{2} + 3 x - 28) = 0$

Factor.

Factor $x^{2} + 3 x - 28$ 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 $- 28$ and whose sum is $3$ .

$- 4, 7$

Write the factored form using these integers.

$- ((x - 4) (x + 7)) = 0$

Remove unnecessary parentheses.

$- (x - 4) (x + 7) = 0$

Step 5

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

$x - 4 = 0$

$x + 7 = 0$

Step 6

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 7

Set $x + 7$ equal to $0$ and solve for $x$ .

Set $x + 7$ equal to $0$ .

$x + 7 = 0$

Subtract $7$ from both sides of the equation.

$x = - 7$

Step 8

The final solution is all the values that make $- (x - 4) (x + 7) = 0$ true.

$x = 4, - 7$

Solve for x 4x^2-28=3x^2-3x

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