The solution for this equation is imaginary solutions or complex solution. To check it, use the formula $b^2-4ac$.
If $b^2-4ac$ is more than 0, then it has different real solution. For instance, x = 2 and x = 3.
If $b^2-4ac$ is equal to 0, then it has same real solution. For instance, x = 3 and x = 3.
If $b^2-4ac$ is less than 0, then it has complex solution. For instance, $x = 2 + 3i$ and $x = 2 - 3i$ where i is referring to imaginary number or $\sqrt{-1}$
In this case, a = 25, b = -30 and c = 58. By using the formula:
$$b^2-4ac=(-30)^2-4(25)(58)$$
$$b^2-4ac=900-5800$$
$$b^2-4ac=-4900 < 0$$
Hence it has complex solution or no real solution (no solution).
So, to solve it, just use Quadratic Formula:
$$x=\frac{b \pm{\sqrt{b^2 - 4ac}}}{2a}$$
$$x=\frac{30 \pm{\sqrt{-4900}}}{50}$$
$$x=\frac{30 \pm{\sqrt{4900}i}}{50}$$
$$x=\frac{30 \pm{70i}}{50}$$
$$x=\frac{3 \pm{7i}}{5}$$
$$x=\frac{3+{7i}}{5},\frac{3-{7i}}{5}$$