### Instruction: The graph of $y=\frac{5x^2-9}{3x^2+5x+2}$ has vertical asymptotes at $x = a$ and $x = b$.  Find $a + b$.### Response: So we're trying to find $a + b$. Let's find $a$ first.
Ok.
To find $a$, we need to find the value of $x$ that makes the denominator of the expression equal to $0$.
And to do that, we need to solve the equation $3x^2 + 5x + 2 = 0$.
Right. Well, the quadratic formula says that the solutions to the equation $ax^2 + bx + c = 0$ are $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
So using that formula, the solutions to $3x^2 + 5x + 2 = 0$ are $x = \frac{-5 \pm \sqrt{5^2 - 4\cdot3\cdot2}}{2\cdot3}$.
So $x = \frac{-5 \pm \sqrt{25 - 24}}{6}$.
Which becomes $x = \frac{-5 \pm 1}{6}$.
So the two solutions are $x = \frac{-5 + 1}{6}$ and $x = \frac{-5 - 1}{6}$.
Which become $x = \frac{-4}{6}$ and $x = \frac{-6}{6}$.
So $x = -\frac{2}{3}$ and $x = -1$.
Ok, so that means that the vertical asymptote is at $x = -\frac{2}{3}$ and the vertical asymptote is at $x = -1$.
Right, so $a + b$ is equal to $-\frac{2}{3} + -1$ which is $-\frac{5}{3}$.