Solve for x 4x^2-36=3x^2-5x

$4 x^{2} - 36 = 3 x^{2} - 5 x$

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

$3 x^{2} - 5 x = 4 x^{2} - 36$

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

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

$3 x^{2} - 5 x - 4 x^{2} = - 36$

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

$- x^{2} - 5 x = - 36$

Add $36$ to both sides of the equation.

$- x^{2} - 5 x + 36 = 0$

Factor the left side of the equation.

Factor $- 1$ out of $- x^{2} - 5 x + 36$ .

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

$- (x^{2}) - 5 x + 36 = 0$

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

$- (x^{2}) - (5 x) + 36 = 0$

Rewrite $36$ as $- 1 (- 36)$ .

$- (x^{2}) - (5 x) - 1 \cdot - 36 = 0$

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

$- (x^{2} + 5 x) - 1 \cdot - 36 = 0$

Factor $- 1$ out of $- (x^{2} + 5 x) - 1 (- 36)$ .

$- (x^{2} + 5 x - 36) = 0$

Factor.

Factor $x^{2} + 5 x - 36$ 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 $- 36$ and whose sum is $5$ .

$- 4, 9$

Write the factored form using these integers.

$- ((x - 4) (x + 9)) = 0$

Remove unnecessary parentheses.

$- (x - 4) (x + 9) = 0$

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

$x - 4 = 0$

$x + 9 = 0$

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 $x + 9$ equal to $0$ and solve for $x$ .

Set $x + 9$ equal to $0$ .

$x + 9 = 0$

Subtract $9$ from both sides of the equation.

$x = - 9$

The final solution is all the values that make $- (x - 4) (x + 9) = 0$ true.

$x = 4, - 9$

Solve for x 4x^2-36=3x^2-5x

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