Solve Using the Quadratic Formula b^2-6b+9=0

$b^{2} - 6 b + 9 = 0$

Step 1

Use the quadratic formula to find the solutions.

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

Step 2

Substitute the values $a = 1$ , $b = - 6$ , and $c = 9$ into the quadratic formula and solve for $b$ .

$\frac{6 \pm \sqrt{{(- 6)}^{2} - 4 \cdot (1 \cdot 9)}}{2 \cdot 1}$

Step 3

Simplify.

Simplify the numerator.

Raise $- 6$ to the power of $2$ .

$b = \frac{6 \pm \sqrt{36 - 4 \cdot 1 \cdot 9}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 9$ .

Multiply $- 4$ by $1$ .

$b = \frac{6 \pm \sqrt{36 - 4 \cdot 9}}{2 \cdot 1}$

Multiply $- 4$ by $9$ .

$b = \frac{6 \pm \sqrt{36 - 36}}{2 \cdot 1}$

Subtract $36$ from $36$ .

$b = \frac{6 \pm \sqrt{0}}{2 \cdot 1}$

Rewrite $0$ as $0^{2}$ .

$b = \frac{6 \pm \sqrt{0^{2}}}{2 \cdot 1}$

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

$b = \frac{6 \pm 0}{2 \cdot 1}$

$6$ plus or minus $0$ is $6$ .

$b = \frac{6}{2 \cdot 1}$

Multiply $2$ by $1$ .

$b = \frac{6}{2}$

Divide $6$ by $2$ .

$b = 3$

Step 4

The final answer is the combination of both solutions.

$b = 3$ Double roots