How do you solve $ {(2x - 3)^2} = 8 $ ?

Hint: First of all we will take square root on both the sides of the equation and then will make the required term “x” the subject and will get the resultant required term value.

Complete step by step solution:
Take the given expression: $ {(2x - 3)^2} = 8 $
Take square-root on both the sides of the equation –
 $ \sqrt {{{(2x - 3)}^2}} = \sqrt 8 $
Square and square-root cancel each other on the left hand side of the equation –
 $ (2x - 3) = 2\sqrt 2 $
Now, move the constant term on the opposite side, when you move any term from one side of the equation to the opposite side then the sign of the term also changes. Positive term become negative and vice-versa.
 $ 2x = 2\sqrt 2 + 3 $
Term multiplication on one side if moved to the opposite side then it goes to the denominator.
 $ x = \dfrac{{2\sqrt 2 + 3}}{2} $
This is the required solution.
So, the correct answer is “ $ x = \dfrac{{2\sqrt 2 + 3}}{2} $ ”.

Note: Always remember the standard equation and formula properly. Follow the given data and conditions carefully. A quadratic equation is an equation of second degree, it means at least one of the terms is squared. Standard equation is $ a{x^2} + bx + c = 0 $ where a,b,and c are constant and “a '' can never be zero and “x” is unknown. Be careful regarding the sign convention. Always remember that the square of negative number or the positive number is always positive. we can find factors by using the formula\[x = \dfrac{{ - b \pm \sqrt \Delta }}{{2a}}\]