Đặt $x^2=a(a\ge 0)$Ta có $pt\iff 10\sqrt{a^3+1}=3(a^2+2)$
$\iff 10(\sqrt{a^3+1}-3a-3)=3a^2+6-30a-30$
$\iff \frac{10(a^3+1-9a^2-18a-9)}{\sqrt{a^3+1}+3a+3}=3a^2-30a-24$
$\iff \frac{10(a^3-9a^2-18a-8)}{\sqrt{a^3+1}+3a+3}=3(a^2-10a-8)$
$\iff \frac{10(a^2-10a-8)(a+1)}{\sqrt{a^3+1}+3a+3}=3(a^2-10a-8)$
$\iff (a^2-10a-8)(\frac{10(a+1)}{\sqrt{a^3+1}+3a+3}-3)=0$
Ta có: $a\ge 0=>\frac{10(a+1)}{\sqrt{a^3+1}+3a+3}-3<0$ (BD tương đương)
Vậy $a^2-10a-8=0\iff a=5+\sqrt{33}(do a\ge 0)=>x=\sqrt{5+\sqrt{33}},x=-\sqrt{5+\sqrt{33}} $