Step 1 :First, we rewrite the equation $6 x^{2}+25 x-25=0$ in the form $(x+D)^{2}=E$ by completing the square. To do this, we first divide the equation by 6 to get $x^{2}+\frac{25}{6}x-\frac{25}{6}=0$.
Step 2 :Next, we take half of the coefficient of $x$, square it, and add it to both sides of the equation. Half of $\frac{25}{6}$ is $\frac{25}{12}$, and its square is $\frac{625}{144}$. Adding this to both sides gives us $x^{2}+\frac{25}{6}x+\frac{625}{144}=\frac{625}{144}+\frac{25}{6}$.
Step 3 :This can be rewritten as $(x+\frac{25}{12})^{2}=\frac{625}{144}+\frac{25}{6}$, which simplifies to $(x+\frac{25}{12})^{2}=\frac{625}{144}+\frac{600}{144}=\frac{1225}{144}$.
Step 4 :Now, we solve for $x$ by taking the square root of both sides. This gives us $x+\frac{25}{12}=\pm\sqrt{\frac{1225}{144}}$, which simplifies to $x+\frac{25}{12}=\pm\frac{35}{12}$.
Step 5 :Subtracting $\frac{25}{12}$ from both sides gives us $x=\pm\frac{35}{12}-\frac{25}{12}$, which simplifies to $x=\frac{10}{12}$ or $x=-\frac{60}{12}$.
Step 6 :Finally, we simplify these fractions to get $x=\frac{5}{6}$ or $x=-5$. So, the solutions to the equation are $x=\frac{5}{6}$ and $x=-5$.
Step 7 :Checking our solutions, we find that they both satisfy the original equation. Therefore, our solutions are correct.
Step 8 :\(\boxed{x=\frac{5}{6}, -5}\)