Answer: \(x=2+\sqrt7\) or \(x=\dfrac23+\dfrac13\sqrt7\).

Use the quadratic formula with

\(a=\sqrt7-2,\qquad b=-4,\qquad c=\sqrt7+2.\)

The discriminant is

\(b^2-4ac=(-4)^2-4(\sqrt7-2)(\sqrt7+2).\)

Now

\((\sqrt7-2)(\sqrt7+2)=7-4=3.\)

So

\(b^2-4ac=16-4(3)=4.\)

Hence

\(x=\frac{4\pm\sqrt4}{2(\sqrt7-2)}.\)

So

\(x=\frac{4\pm2}{2(\sqrt7-2)}.\)

The first root is

\(x=\frac6{2(\sqrt7-2)} =\frac3{\sqrt7-2}.\)

Rationalise the denominator:

\(\frac3{\sqrt7-2}\cdot\frac{\sqrt7+2}{\sqrt7+2} =\frac{3(\sqrt7+2)}{7-4} =\sqrt7+2.\)

The second root is

\(x=\frac2{2(\sqrt7-2)} =\frac1{\sqrt7-2}.\)

Rationalise the denominator:

\(\frac1{\sqrt7-2}\cdot\frac{\sqrt7+2}{\sqrt7+2} =\frac{\sqrt7+2}{3}.\)

Therefore

\(x=2+\sqrt7\quad\text{or}\quad x=\frac23+\frac13\sqrt7.\)