Does the equation $5{\sqrt{6x^2-19x-20}}+\frac{1}{4}{\sqrt{4(x^2+1)-17x}}=0$ have any real solutions?

Question

For all value of $x$, ${\sqrt{6x^2-19x-20}}>0$ and ${\sqrt{4(x^2+1)-17x}}>0$. Does the equation $5{\sqrt{6x^2-19x-20}}+\frac{1}{4}{\sqrt{4(x^2+1)-17x}}=0$ have any real solutions? Any how to obtain the solutions?

Answer 1

Notice that $6x^2-19x-20=(x-4)(6x+5)$ and $4(x^2+1)-17x=(x-4)(4x-1)$.

Hence $x=4$ is the only real solution.