To solve this equation, let's make a substitution to simplify the equation. Let [tex]y = \sqrt[4]{x}[/tex]. Then, we have:

[tex]\sqrt{x} - 4y + 3 = 0[/tex]

Now, let's express [tex]\sqrt{x}[/tex] in terms of y:

[tex]\sqrt{x} = y^2[/tex]

Substitute this back into the equation:

[tex]y^2 - 4y + 3 = 0[/tex]

This is now a quadratic equation that can be factored easily:

tex(y-3) = 0[/tex]

So, [tex]y = 1[/tex] or [tex]y = 3[/tex]

Now, substitute back [tex]y = \sqrt[4]{x}[/tex] and solve for x:

For [tex]y = 1[/tex]:
[tex]\sqrt[4]{x} = 1[/tex]
[tex]x = 1^4[/tex]
[tex]x = 1[/tex]

For [tex]y = 3[/tex]:
[tex]\sqrt[4]{x} = 3[/tex]
[tex]x = 3^4[/tex]
[tex]x = 81[/tex]

Therefore, the solutions to the equation are [tex]x = 1[/tex] and [tex]x = 81[/tex].