Solve by Factoring (7x+3)(2x-6)=0

$(7 x + 3) (2 x - 6) = 0$

Step 1

Factor.

Factor $2$ out of $2 x - 6$ .

Factor $2$ out of $2 x$ .

$(7 x + 3) (2 (x) - 6) = 0$

Factor $2$ out of $- 6$ .

$(7 x + 3) (2 x + 2 \cdot - 3) = 0$

Factor $2$ out of $2 x + 2 \cdot - 3$ .

$(7 x + 3) (2 (x - 3)) = 0$

Remove unnecessary parentheses.

$(7 x + 3) \cdot (2 (x - 3)) = 0$

Step 2

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

$7 x + 3 = 0$

$x - 3 = 0$

Step 3

Set $7 x + 3$ equal to $0$ and solve for $x$ .

Set $7 x + 3$ equal to $0$ .

$7 x + 3 = 0$

Solve $7 x + 3 = 0$ for $x$ .

Subtract $3$ from both sides of the equation.

$7 x = - 3$

Divide each term in $7 x = - 3$ by $7$ and simplify.

Divide each term in $7 x = - 3$ by $7$ .

$\frac{7 x}{7} = \frac{- 3}{7}$

Simplify the left side.

Cancel the common factor of $7$ .

Cancel the common factor.

$\frac{7 x}{7} = \frac{- 3}{7}$

Divide $x$ by $1$ .

$x = \frac{- 3}{7}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{3}{7}$

Step 4

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.

$x = 3$

Step 5

The final solution is all the values that make $(7 x + 3) \cdot (2 (x - 3)) = 0$ true.

$x = - \frac{3}{7}, 3$