I will assume that $a$ and $b$ are $\ge 0$, but are not both $0$. Take the square root(s). We get $\sqrt{a} \,x=\pm\sqrt{b}(x-10).$ Now we have two linear equations. Deal with them separately.
The equation $\sqrt{a}x=\sqrt{b}(x-10)$ can be rewritten as $\sqrt{a}\,x =\sqrt{b}\, x-10\sqrt{b}$ and then as $10\sqrt{b}=(\sqrt{b}-\sqrt{a})x$. If $a\ne b$, it has solution $x=\frac{10\sqrt{b}}{\sqrt{b}-\sqrt{a}} .$ If $a=b$ the linear equation has no solution. The other equation is dealt with similarly. It gives after a while $x=\frac{10\sqrt{b}}{\sqrt{a}+\sqrt{b}} .$
Remark: As written, the equation was awfully close to a linear (that is, nice) equation. If I am that close, I prefer to go directly for the prize.
For completeness, we deal with the other possibilities for $a$ and $b$. If $a$ and $b$ are both $\le 0$, but are not both $0$, use the equivalent equation $|a|x^2=|b|(x-10)^2$. If $a$ is positive and $b$ is negative, or the other way around, there is no solution. And if $a$ and $b$ are both $0$, then every real number $x$ is a solution.