Помогите решить два иррационального уравнения вообще не шарю

Решите задачу:

$1)\quad \sqrt{5+\sqrt{x-4}}=3\; ,\; \; \; ODZ:\; \left \{ {{x \geq 4} \atop {5+\sqrt{x-4} \geq 0}} \right. \; \to \; x \geq 4\\\\5+\sqrt{x-4}=9\\\\\sqrt{x-4}=4\\\\x-4=16\\\\x=20\\\\Proverka;\; \; \sqrt{5+\sqrt{20-4}}=3\; ,\; \; \sqrt{5+4}=3\; ,\; 3=3\\\\Otvet:\; \; x=20\; .$

$2)\quad \sqrt{2-\sqrt{2+x}}=x\; ,\\\\ ODZ:\; \left \{ {{x \geq 0,\; x+2 \geq 0} \atop {2-\sqrt{x+2} \geq 0}} \right. \; \left \{ {{x \geq 0} \atop {\sqrt{x+2}\leq 2}} \right. \; \left \{ {{x \geq 0} \atop {x+2 \leq 4}} \right. \; \left \{ {{x \geq 0} \atop {x \leq 2}} \right. \; \; \; 0 \leq x \leq 2\\\\2-\sqrt{2+x}=x^2\\\\\sqrt{2+x}=2-x^2\\\\2+x=(2-x^2)^2\\\\2+x=4-4x^2+x^4\\\\x^4-4x^2-x+2=0\\\\(x^4+x^3)-x^3-x^2-3x^2-x+2=0\\\\x^3(x+1)-x^2(x+1)-(3x^2+x-2)=0\\\\x^3(x+1)-x^2(x+1)-3(x+1)(x-\frac{2}{3})=0$

$(x+1)(x^3-x^2-3x+2)=0\\\\x_1=-1\notin ODZ\\\\x^3-x^2-3x+2=0\\\\x^2(x-2)+x(x-2)-(x-2)=0\\\\(x-2)(x^2+x-1)=0\; \; \to \; \; x_2=2\in ODZ\; ,\\\\x^2+x-1=0\\\\D=1+5=6\; ,\; \; x_{3,4}=\frac{-1\pm \sqrt6}{2}= \left [ {{\approx -1,72\notin ODZ} \atop {\approx 0,72\in ODZ}} \right. \\\\Otvet:\; \; x=2\; ,\; x=\frac{-1+\sqrt6}{2}\; .$