\( \begin{align} {\left( {2x - 3} \right)^2} &= 1\\ \left( {2x - 3} \right)\left( {2x - 3} \right) &= 1\\ 4{x^2} - 12x + 9 &= 1\\ 4{x^2} - 12x + 8 &= 0\\ 4\left( {x - 1} \right)\left( {x - 2} \right) &= 0\\ \,x &= 1\,of\,x = 2 \end{align} \)

\( \begin{align} {\left( {2x - 3} \right)^2} &= 1\\ \left( {2x - 3} \right)\left( {2x - 3} \right) &= 1\\ 4{x^2} - 12x + 9 &= 1\\ 4{x^2} - 12x + 8 &= 0\\ x &= \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\\ x& = \frac{{ - \left( { - 12} \right) \pm \sqrt {{{\left( { - 12} \right)}^2} - 4\left( 4 \right)\left( 8 \right)} }}{{2\left( 4 \right)}}\\ x &= 1\,\,\,of\,\,\,x = 2 \end{align} \)