From Förberedande kurs i matematik 1

First, we move the 2 to the right-hand side to get \displaystyle \sqrt{3x-8}=x-2 and then square away the root sign,

\displaystyle 3x-8 = (x-2)^{2}

(*)

or, with the right-hand side expanded

\displaystyle 3x-8=x^{2}-4x+4\,\textrm{.}

If we move over all the terms to the left-hand side, we get

\displaystyle x^{2}-7x+12=0\,\textrm{.}

If we complete the square of the left-hand side,

\displaystyle \begin{align} x^2-7x+12 &= \Bigl(x-\frac{7}{2}\Bigr)^2 - \Bigl(\frac{7}{2}\Bigr)^2 + 12\\[5pt] &= \Bigl(x-\frac{7}{2}\Bigr)^2 - \frac{49}{4} + \frac{48}{4}\\[5pt] &= \Bigl(x-\frac{7}{2}\Bigr)^2 - \frac{1}{4} \end{align}

the equation can be written as

\displaystyle \Bigl(x-\frac{7}{2}\Bigr)^{2} = \frac{1}{4}

and the solutions are

• \displaystyle x = \frac{7}{2} + \sqrt{\frac{1}{4}} = \frac{7}{2} + \frac{1}{2} = \frac{8}{2} = 4\,,

• \displaystyle x = \frac{7}{2} - \sqrt{\frac{1}{4}} = \frac{7}{2} - \frac{1}{2} = \frac{6}{2} = 3\,\textrm{.}

To be on the safe side, we verify that \displaystyle x=3 and \displaystyle x=4 satisfy the squared equation (*)

x = 3:	\displaystyle \ \text{LHS} = 3\cdot 3-8 = 9-8 = 1 and
	\displaystyle \ \text{RHS} = (3-2)^2 = 1
x = 4:	\displaystyle \ \text{LHS} = 3\cdot 4-8 = 12-8 = 4 and
	\displaystyle \ \text{RHS} = (4-2)^2 = 4

Because we squared the root equation, possible spurious roots turn up and we therefore have to verify the solutions when we go back to the original root equation:

x = 3:	\displaystyle \ \text{LHS} = \sqrt{3\cdot 3-8} + 2 = \sqrt{9-8} + 2 = 1+2 = 3 and
	\displaystyle \ \text{RHS} = 3
x = 4:	\displaystyle \ \text{LHS} = \sqrt{3\cdot 4-8}-2 = \sqrt{12-8}+2 = 2+2 = 4 and
	\displaystyle \ \text{RHS} = 4

The solutions to the root equation are \displaystyle x=3 and \displaystyle x=4.