I have the following equation to solve. I know that the answer is -5, I made several attempts at this, and arrive at a different answer. My first thought was to factor out the trinomial, but that didn't help. What are the correct steps to solve this equation for x?
$$\frac{x^2 + x -2}{x+3} ={-9}$$
On
I can't comment the other answer so just elaborating on each step:
$$\frac{x^2 + x -2}{x+3} ={-9}$$
Multiply both sides of the equation by $x + 3$ $$\frac{x^2 + x -2}{x+3} * \frac{x+3}{1} ={-9} * \frac{x + 3}{1}$$
$$x^2+x-2=-9x-27$$
Then subtract $-9x -27$ or alternatively add $9x + 27$ to both sides.
$$x^2+10x +25 = 0$$
Then, use quadratic formula or any other method you know.
$$\frac{x^2 + x -2}{x+3} ={-9}$$ $$x^2+x-2=-9x-27$$ $$x^2+10x +25 = 0$$
now use the quadratic equation $x = \dfrac{-B \pm \sqrt{B^2-4AC}}{2A}$
and you get the result $x=-5$