Solve Using the Quadratic Formula (2x-1)^2-10(2x-1)+9=0

${(2 x - 1)}^{2} - 10 (2 x - 1) + 9 = 0$

Step 1

Simplify the left side.

Simplify ${(2 x - 1)}^{2} - 10 (2 x - 1) + 9$ .

Simplify each term.

Apply the distributive property.

${(2 x - 1)}^{2} - 10 (2 x) - 10 \cdot - 1 + 9 = 0$

Multiply $2$ by $- 10$ .

${(2 x - 1)}^{2} - 20 x - 10 \cdot - 1 + 9 = 0$

Multiply $- 10$ by $- 1$ .

${(2 x - 1)}^{2} - 20 x + 10 + 9 = 0$

Add $10$ and $9$ .

${(2 x - 1)}^{2} - 20 x + 19 = 0$

Step 2

Use the quadratic formula to find the solutions.

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

Step 3

Substitute the values $a = 4$ , $b = - 24$ , and $c = 20$ into the quadratic formula and solve for $x$ .

$\frac{24 \pm \sqrt{{(- 24)}^{2} - 4 \cdot (4 \cdot 20)}}{2 \cdot 4}$

Step 4

Simplify.

Simplify the numerator.

Raise $- 24$ to the power of $2$ .

$x = \frac{24 \pm \sqrt{576 - 4 \cdot 4 \cdot 20}}{2 \cdot 4}$

Multiply $- 4 \cdot 4 \cdot 20$ .

Multiply $- 4$ by $4$ .

$x = \frac{24 \pm \sqrt{576 - 16 \cdot 20}}{2 \cdot 4}$

Multiply $- 16$ by $20$ .

$x = \frac{24 \pm \sqrt{576 - 320}}{2 \cdot 4}$

Subtract $320$ from $576$ .

$x = \frac{24 \pm \sqrt{256}}{2 \cdot 4}$

Rewrite $256$ as $16^{2}$ .

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

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

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

Multiply $2$ by $4$ .

$x = \frac{24 \pm 16}{8}$

Simplify $\frac{24 \pm 16}{8}$ .

$x = 3 \pm 2$

Step 5

The final answer is the combination of both solutions.

$x = 5, 1$