Calculate: (3 x+2) (x-1) = 7-7 x

Expression: $$( 3 x + 2 ) ( x - 1 ) = 7 - 7 x$$

Use the distributive property to multiply $3x+2$ by $x-1$ and combine like terms.

$$3x^{2}-x-2=7-7x$$

Add $7x$ to both sides.

$$3x^{2}-x-2+7x=7$$

Combine $-x$ and $7x$ to get $6x$.

$$3x^{2}+6x-2=7$$

Add $2$ to both sides.

$$3x^{2}+6x=7+2$$

Add $7$ and $2$ to get $9$.

$$3x^{2}+6x=9$$

Divide both sides by $3$.

$$\frac{3x^{2}+6x}{3}=\frac{9}{3}$$

Dividing by $3$ undoes the multiplication by $3$.

$$x^{2}+\frac{6}{3}x=\frac{9}{3}$$

Divide $6$ by $3$.

$$x^{2}+2x=\frac{9}{3}$$

Divide $9$ by $3$.

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

Divide $2$, the coefficient of the $x$ term, by $2$ to get $1$. Then add the square of $1$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+2x+1^{2}=3+1^{2}$$

Square $1$.

$$x^{2}+2x+1=3+1$$

Add $3$ to $1$.

$$x^{2}+2x+1=4$$

Factor $x^{2}+2x+1$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x+1\right)^{2}=4$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{4}$$

Simplify.

$$x+1=2$$ $$x+1=-2$$

Subtract $1$ from both sides of the equation.

$$x=1$$ $$x=-3$$