Proper method for solving quadratic equations with exponents $(\sqrt {x^2-5x+6}+\sqrt{x^2-5x+4})^{x/2}$ + $(\sqrt {x^2-5x+6}-\sqrt{x^2-5x+4})^{x/2}$ = $2^{(x+4)/4}$
I have found out, by trial and error method, that $x=0$ and $x=4$ satisfy this equation. But is there a proper way to solve this equation and get the solutions?

Put
$$
a=(\sqrt {x^2-5x+6}+\sqrt{x^2-5x+4})^{x/2},\\
b=(\sqrt {x^2-5x+6}-\sqrt{x^2-5x+4})^{x/2}
$$
Then 
$$
a+b=2^{(x+4)/4},\\
a \cdot b=2^{x/2}
$$
By solving it we get that $a=b,$ or
$$
(\sqrt {x^2-5x+6}+\sqrt{x^2-5x+4})^{x/2}=(\sqrt {x^2-5x+6}-\sqrt{x^2-5x+4})^{x/2}.
$$
It is a simple equation.
