HELP PLEASE

Solve the following equation!

$\begin{array}{rcl} \sqrt{x-4} -\sqrt[4]{x-4} &=& 12 \\ \end{array}$

The solution(s) of the given equation is/are x=??

$\begin{array}{rcll} \sqrt{x-4} -\sqrt[4]{x-4} &=& 12 \qquad &| \qquad +\sqrt[4]{x-4}\\ \sqrt{x-4} &=& 12 + \sqrt[4]{x-4} \qquad &| \qquad -12\\ \sqrt{x-4}-12 &=& \sqrt[4]{x-4} \qquad &| \qquad()^2\\ ( \sqrt{x-4}-12 )^2 &=& ( \sqrt[4]{x-4} )^2 \\ x-4 - 24\sqrt{x-4} + 144 &=& (x-4)^{\frac24 } \\ x-4 - 24\sqrt{x-4} + 144 &=& (x-4)^{\frac12 } \\ x-4 - 24\sqrt{x-4} + 144 &=& \sqrt{x-4} \qquad &| \qquad + 24\sqrt{x-4}\\ x-4 + 144 &=& \sqrt{x-4}+ 24\sqrt{x-4}\\ x-4 + 144 &=& 25\sqrt{x-4}\\ x + 140 &=& 25\sqrt{x-4}\qquad &| \qquad()^2\\ ( x + 140 )^2 &=& ( 25\sqrt{x-4} )^2 \\ x^2+280x+140^2 &=& 625 ( x-4)\\ x^2+280x+19600 &=& 625x-2500 \qquad &| \qquad -625x\\ x^2-345x +19600 &=& -2500 \qquad &| \qquad +2500\\ x^2-345x +22100 &=& 0 \\ \end{array}$

$\begin{array}{rcll} x^2-345x +22000 &=& 0\\ \qquad \boxed{ ~ x = {-b \pm \sqrt{b^2-4ac} \over 2a} ~}\\ x &=& {345 \pm \sqrt{345^2-4\cdot22100} \over 2}\\ x &=& {345 \pm \sqrt{30625} \over 2}\\ x &=& {345 \pm 175 \over 2}\\ \hline x_1 &=& {345 + 175 \over 2}\\ \mathbf{x_1} &\mathbf{=} & \mathbf{260}\\ \hline x_2 &=& {345 - 175 \over 2}\\ x_2 &=& 85 \quad \text{ no solution}\\ \end{array}$

The solution is x = 260