Решите уравнение : корень из (х^2 - 16 )= х^2-22

$\sqrt{(x^2-16)}=x^2-22 \\ (x^2-16)=(x^2-22)^2 \\ x^2-16=x^4-44x^2+484 \\ x^4-45x^2+500=0 \\ x^2=t \\ t^2-45t+500=0 \\ \\ D=b^2-4ac=(-45)^2-4*1*500=2025-2000=25 \\ \\ t_{1}= \frac{-b+ \sqrt{D} }{2a} = \frac{45+5}{2} =25 \\ \\ t_{2}= \frac{-b- \sqrt{D} }{2a} = \frac{45-5}{2} =20 \\ \\ x^2=25, x_1=5,x_2=-5 \\ \\ x^2=20, x_3=2 \sqrt{5},x_4=-2 \sqrt{5}$

проверка:
$\sqrt{(5^2-16)}=5^2-22 \\ \sqrt{(25-16)}=25-22 \\ \sqrt{9}=3 \\ \\ \sqrt{((-5)^2-16)}=(-5)^2-22 \\ \sqrt{(25-16)}=25-22 \\ \sqrt{9}=3 \\ \\ \sqrt{((2\sqrt{5})^2-16)}=(2\sqrt{5})^2-22 \\ \sqrt{((\sqrt{20})^2-16)}=(\sqrt{20})^2-22 \\ \sqrt{(20-16)}=20-22 \\ \sqrt{4}=2 \\ \\ \sqrt{((-2\sqrt{5})^2-16)}=(-2\sqrt{5})^2-22 \\ \sqrt{((-\sqrt{20})^2-16)}=(-\sqrt{20})^2-22 \\ \sqrt{(20-16)}=20-22 \\ \sqrt{4}=2$