We are asked to find the roots of the quadratic equation $x^2 + \frac{21}{4}x = -\frac{9}{4}$.

1. Problem Description

We are asked to find the roots of the quadratic equation $x^2 + \frac{21}{4}x = -\frac{9}{4}$.

2. Solution Steps

First, we rewrite the equation in the standard quadratic form $ax^2 + bx + c = 0$.

$x^2 + \frac{21}{4}x + \frac{9}{4} = 0$

To eliminate the fractions, multiply both sides of the equation by 4:

$4(x^2 + \frac{21}{4}x + \frac{9}{4}) = 4(0)$

$4x^2 + 21x + 9 = 0$

Now we can try to factor the quadratic expression. We are looking for two numbers that multiply to $4 \cdot 9 = 36$ and add up to $21$. These numbers are $3$ and $18$.

Rewrite the middle term:

$4x^2 + 3x + 18x + 9 = 0$

Factor by grouping:

$x(4x + 3) + \frac{9}{2}(4x + 3) = 0$ or $3x + 18x= 21x$

$x(4x + 3) + \frac{9}{2}(4x+3)$ not right way

We can also factor as follows:

$4x^2 + 21x + 9 = (4x + 3)(x + 3) = 0$.

Set each factor equal to zero:

$4x + 3 = 0$ or $x + 3 = 0$

Solve for $x$:

$4x = -3$, so $x = -\frac{3}{4}$

$x = -3$

Therefore, the roots are $x = -\frac{3}{4}$ and $x = -3$.

3. Final Answer

-3/4, -3