Đặt $\sqrt x=a(a\ge 0)$. Phương trình trở thành:
$\begin{array}{l}
{a^4} - 7{a^2} - 4a + 20 = 0\\
\Leftrightarrow {a^4} - 2{a^3} + 2{a^3} - 4{a^2} - 3{a^2} + 6a - 10a + 20 = 0\\
\Leftrightarrow {a^3}\left( {a - 2} \right) + 2{a^2}\left( {a - 2} \right) - 3a\left( {a - 2} \right) - 10\left( {a - 2} \right) = 0\\
\Leftrightarrow \left( {a - 2} \right)\left( {{a^3} + 2{a^2} - 3a - 10} \right) = 0\\
\Leftrightarrow \left( {a - 2} \right)\left[ {{a^3} - 2{a^2} + 4{a^2} - 8a + 5a - 10} \right] = 0\\
\Leftrightarrow {\left( {a - 2} \right)^2}\underbrace {\left( {{a^2} + 4a + 5} \right)}_{ > 0} = 0\\
\Leftrightarrow a = 2 \Leftrightarrow \sqrt x = 2 \Leftrightarrow x = 4
\end{array}$
