We'll have to repeatedly square both sides of the equation, in order to get rid of the square roots. Squaring a first time yields
[tex]19+\sqrt{30+\sqrt{32+x}}=25[/tex]
Move the 19 to the right hand side:
[tex]\sqrt{30+\sqrt{32+x}}=6[/tex]
And square again:
[tex]30+\sqrt{32+x}=36 \iff \sqrt{32+x}=6[/tex]
Square one last time:
[tex]32+x=36 \iff x=36-32=4[/tex]
Let's check the solutions: all these squaring might have created external solutions:
[tex]\sqrt{19+\sqrt{30+\sqrt{32+4}}}=\sqrt{19+\sqrt{30+6}}=\sqrt{19+6}=\sqrt{25}=5[/tex]
So, [tex]x=4[/tex] is a feasible solution.
