quadratic pre colloge problem

Question

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 solution of $x^2-3x+c=0$ are....

i cant do this problem due to language barrier

Answer 1

Suppose $x_1$ and $x_2$ are roots of $x^2-3x+c$, and $-x_1$ and $x_3$ are roots of $x^2+3x-c$, then we know that $$x_1^2-3x_1+c=0=(-x_1)^2+3(-x_1)-c$$

so $c=?$. Then the roots of both equations become clear (hopefully).