From Förberedande kurs i matematik 1

We divide both sides by \displaystyle 3 and complete the square on the left-hand side:

\displaystyle \begin{align} & x^{2}-\frac{10}{3}x+\frac{8}{3}=\left( x-\frac{5}{3} \right)^{2}-\left( -\frac{5}{3} \right)^{2}+\frac{8}{3}=\left( x-\frac{5}{3} \right)^{2}-\frac{25}{9}+\frac{24}{9} \\ & =\left( x-\frac{5}{3} \right)^{2}-\frac{1}{9} \\ \end{align}

The equation then becomes

\displaystyle \left( x-\frac{5}{3} \right)^{2}=\frac{1}{9}

and taking the root gives the solutions as

\displaystyle x-\frac{5}{3}=\sqrt{\frac{1}{9}}=\frac{1}{3} i.e. \displaystyle x=\frac{5}{3}+\frac{1}{3}=\frac{6}{3}=2.

\displaystyle x-\frac{5}{3}=-\sqrt{\frac{1}{9}}=-\frac{1}{3} i.e. \displaystyle x=\frac{5}{3}-\frac{1}{3}=\frac{4}{3}.

Check:

\displaystyle x=\text{4}/\text{3}: LHS \displaystyle =3\centerdot \left( \frac{4}{3} \right)^{2}-10\centerdot \frac{4}{3}+8=3\centerdot \frac{16}{9}-\frac{40}{3}+\frac{8\centerdot 3}{3}=\frac{16-40+24}{3}=0= RHS

\displaystyle x=\text{2}: LHS \displaystyle =3\centerdot 2^{2}-10\centerdot 2+8=12-20+8=0= RHS