If $c$ is a real number and the negative of one of the solutions of $x^2-3x+c=0$ is a solution of $x^2+3x-c=0$, then the solutions of $x^2-3x+c=0$ are

$\textbf{(A) }1,~2\qquad\textbf{(B) }-1,~-2\qquad\textbf{(C) }0,~3\qquad\textbf{(D) }0,~-3\qquad \textbf{(E) }\frac{3}{2},~\frac{3}{2}$
1. Let the solutions of the quadratic equation \(x^2 - 3x + c = 0\) be \( \alpha \) and \( \beta \). By Vieta's formulas, we know:
   \[
   \alpha + \beta = 3 \quad \text{and} \quad \alpha \beta = c
   \]

2. According to the problem, the negative of one of the solutions of \(x^2 - 3x + c = 0\) is a solution of \(x^2 + 3x - c = 0\). Without loss of generality, assume \(-\alpha\) is a solution of \(x^2 + 3x - c = 0\). Then, substituting \(-\alpha\) into the second equation, we get:
   \[
   (-\alpha)^2 + 3(-\alpha) - c = 0
   \]
   Simplifying, we have:
   \[
   \alpha^2 - 3\alpha - c = 0
   \]

3. Since \(\alpha\) is a solution of \(x^2 - 3x + c = 0\), we also have:
   \[
   \alpha^2 - 3\alpha + c = 0
   \]

4. Equating the two expressions for \(\alpha^2 - 3\alpha\), we get:
   \[
   \alpha^2 - 3\alpha - c = \alpha^2 - 3\alpha + c
   \]
   Subtracting \(\alpha^2 - 3\alpha\) from both sides, we obtain:
   \[
   -c = c
   \]
   This implies:
   \[
   c = 0
   \]

5. Substituting \(c = 0\) back into the original equation \(x^2 - 3x + c = 0\), we get:
   \[
   x^2 - 3x = 0
   \]
   Factoring, we have:
   \[
   x(x - 3) = 0
   \]
   Therefore, the solutions are:
   \[
   x = 0 \quad \text{or} \quad x = 3
   \]

Conclusion:
The solutions of \(x^2 - 3x + c = 0\) are \(\boxed{0, 3}\).