To find an equation that has the solution set \( x = \left{ -\frac{9}{5}, 3 \right} \), we can create a quadratic equation. The solutions or roots of a quadratic equation \( ax^2 + bx + c = 0 \) can be represented as \( (x - r_1)(x - r_2) = 0 \), where \( r_1 \) and \( r_2 \) are the roots.
In this case, the roots are:
- \( r_1 = -\frac{9}{5} \)
- \( r_2 = 3 \)
We can write the equation as:
\[ \left(x + \frac{9}{5}\right)(x - 3) = 0 \]
To eliminate the fraction, multiply through by 5:
\[ 5\left(x + \frac{9}{5}\right)(x - 3) = 0 \implies (5x + 9)(x - 3) = 0 \]
Now expand this:
\[ (5x + 9)(x - 3) = 5x^2 - 15x + 9x - 27 = 5x^2 - 6x - 27 \]
Thus, the equation we are looking for is:
\[ 5x^2 - 6x - 27 = 0 \]
This equation has the solutions \( x = -\frac{9}{5} \) and \( x = 3 \). Therefore, the final answer is:
\[ 5x^2 - 6x - 27 = 0 \]