It looks like you have an equation that needs to be solved. However, the equation you provided is a bit unclear due to the formatting. I'll assume you meant to write:

\[ \frac{3x^2 + 7x - 10}{x} = 0 \]

To solve this equation, follow these steps:

1. **Simplify the equation**: Since the denominator \( x \) cannot be zero, we can multiply both sides of the equation by \( x \) to eliminate the fraction:

\[ 3x^2 + 7x - 10 = 0 \]

2. **Solve the quadratic equation**: This is a standard quadratic equation of the form \( ax^2 + bx + c = 0 \). We can solve it using the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Here, \( a = 3 \), \( b = 7 \), and \( c = -10 \).

3. **Calculate the discriminant**: The discriminant \( \Delta \) is given by \( b^2 - 4ac \):

\[ \Delta = 7^2 - 4(3)(-10) \]
\[ \Delta = 49 + 120 \]
\[ \Delta = 169 \]

4. **Find the roots**: Now, use the quadratic formula to find the roots:

\[ x = \frac{-7 \pm \sqrt{169}}{2 \cdot 3} \]
\[ x = \frac{-7 \pm 13}{6} \]

This gives us two solutions:

\[ x_1 = \frac{-7 + 13}{6} = \frac{6}{6} = 1 \]

\[ x_2 = \frac{-7 - 13}{6} = \frac{-20}{6} = -\frac{10}{3} \]

So, the solutions to the equation are:

\[ x = 1 \quad \text{or} \quad x = -\frac{10}{3} \]

Remember that \( x \) cannot be zero because it was in the denominator of the original equation. Therefore, the solutions \( x = 1 \) and \( x = -\frac{10}{3} \) are valid.