Solve Using the Square Root Property (2x-3)^2=39

${(2 x - 3)}^{2} = 39$

Step 1

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$2 x - 3 = \pm \sqrt{39}$

Step 2

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$2 x - 3 = \sqrt{39}$

Add $3$ to both sides of the equation.

$2 x = \sqrt{39} + 3$

Divide each term in $2 x = \sqrt{39} + 3$ by $2$ and simplify.

Divide each term in $2 x = \sqrt{39} + 3$ by $2$ .

$\frac{2 x}{2} = \frac{\sqrt{39}}{2} + \frac{3}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{\sqrt{39}}{2} + \frac{3}{2}$

Divide $x$ by $1$ .

$x = \frac{\sqrt{39}}{2} + \frac{3}{2}$

Next, use the negative value of the $\pm$ to find the second solution.

$2 x - 3 = - \sqrt{39}$

Add $3$ to both sides of the equation.

$2 x = - \sqrt{39} + 3$

Divide each term in $2 x = - \sqrt{39} + 3$ by $2$ and simplify.

Divide each term in $2 x = - \sqrt{39} + 3$ by $2$ .

$\frac{2 x}{2} = \frac{- \sqrt{39}}{2} + \frac{3}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{- \sqrt{39}}{2} + \frac{3}{2}$

Divide $x$ by $1$ .

$x = \frac{- \sqrt{39}}{2} + \frac{3}{2}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{\sqrt{39}}{2} + \frac{3}{2}$

The complete solution is the result of both the positive and negative portions of the solution.

$x = \frac{\sqrt{39}}{2} + \frac{3}{2}, - \frac{\sqrt{39}}{2} + \frac{3}{2}$

Step 3

The result can be shown in multiple forms.

Exact Form:

$x = \frac{\sqrt{39}}{2} + \frac{3}{2}, - \frac{\sqrt{39}}{2} + \frac{3}{2}$

Decimal Form:

$x = 4.62249899 \dots, - 1.62249899 \dots$