Find All Complex Solutions (2x-1)^2=25

${(2 x - 1)}^{2} = 25$

Move all terms to the left side of the equation and simplify.

Subtract $25$ from both sides of the equation.

${(2 x - 1)}^{2} - 25 = 0$

Simplify ${(2 x - 1)}^{2} - 25$ .

Simplify each term.

Rewrite ${(2 x - 1)}^{2}$ as $(2 x - 1) (2 x - 1)$ .

$(2 x - 1) (2 x - 1) - 25 = 0$

Expand $(2 x - 1) (2 x - 1)$ using the FOIL Method.

Apply the distributive property.

$2 x (2 x - 1) - 1 (2 x - 1) - 25 = 0$

Apply the distributive property.

$2 x (2 x) + 2 x \cdot - 1 - 1 (2 x - 1) - 25 = 0$

Apply the distributive property.

$2 x (2 x) + 2 x \cdot - 1 - 1 (2 x) - 1 \cdot - 1 - 25 = 0$

Simplify and combine like terms.

Simplify each term.

Rewrite using the commutative property of multiplication.

$2 \cdot 2 x \cdot x + 2 x \cdot - 1 - 1 (2 x) - 1 \cdot - 1 - 25 = 0$

Multiply $x$ by $x$ by adding the exponents.

Move $x$ .

$2 \cdot 2 (x \cdot x) + 2 x \cdot - 1 - 1 (2 x) - 1 \cdot - 1 - 25 = 0$

Multiply $x$ by $x$ .

$2 \cdot 2 x^{2} + 2 x \cdot - 1 - 1 (2 x) - 1 \cdot - 1 - 25 = 0$

Multiply $2$ by $2$ .

$4 x^{2} + 2 x \cdot - 1 - 1 (2 x) - 1 \cdot - 1 - 25 = 0$

Multiply $- 1$ by $2$ .

$4 x^{2} - 2 x - 1 (2 x) - 1 \cdot - 1 - 25 = 0$

Multiply $2$ by $- 1$ .

$4 x^{2} - 2 x - 2 x - 1 \cdot - 1 - 25 = 0$

Multiply $- 1$ by $- 1$ .

$4 x^{2} - 2 x - 2 x + 1 - 25 = 0$

Subtract $2 x$ from $- 2 x$ .

$4 x^{2} - 4 x + 1 - 25 = 0$

Subtract $25$ from $1$ .

$4 x^{2} - 4 x - 24 = 0$

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Substitute the values $a = 4$ , $b = - 4$ , and $c = - 24$ into the quadratic formula and solve for $x$ .

$\frac{4 \pm \sqrt{{(- 4)}^{2} - 4 \cdot (4 \cdot - 24)}}{2 \cdot 4}$

Simplify.

Simplify the numerator.

Raise $- 4$ to the power of $2$ .

$x = \frac{4 \pm \sqrt{16 - 4 \cdot 4 \cdot - 24}}{2 \cdot 4}$

Multiply $- 4$ by $4$ .

$x = \frac{4 \pm \sqrt{16 - 16 \cdot - 24}}{2 \cdot 4}$

Multiply $- 16$ by $- 24$ .

$x = \frac{4 \pm \sqrt{16 + 384}}{2 \cdot 4}$

Add $16$ and $384$ .

$x = \frac{4 \pm \sqrt{400}}{2 \cdot 4}$

Rewrite $400$ as $20^{2}$ .

$x = \frac{4 \pm \sqrt{20^{2}}}{2 \cdot 4}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \frac{4 \pm 20}{2 \cdot 4}$

Multiply $2$ by $4$ .

$x = \frac{4 \pm 20}{8}$

Simplify $\frac{4 \pm 20}{8}$ .

$x = \frac{1 \pm 5}{2}$

The final answer is the combination of both solutions.

$x = 3, - 2$

Find All Complex Solutions (2x-1)^2=25

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