$$3x^{2}-x-2=7-7x$$
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.$$x=1$$ $$x=-3$$