Solve Using the Quadratic Formula r^2-10r+25=0

$r^{2} - 10 r + 25 = 0$

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 $a = 1$ , $b = - 10$ , and $c = 25$ into the quadratic formula and solve for $r$ .

$\frac{10 \pm \sqrt{{(- 10)}^{2} - 4 \cdot (1 \cdot 25)}}{2 \cdot 1}$

Step 3

Simplify.

Simplify the numerator.

Raise $- 10$ to the power of $2$ .

$r = \frac{10 \pm \sqrt{100 - 4 \cdot 1 \cdot 25}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 25$ .

Multiply $- 4$ by $1$ .

$r = \frac{10 \pm \sqrt{100 - 4 \cdot 25}}{2 \cdot 1}$

Multiply $- 4$ by $25$ .

$r = \frac{10 \pm \sqrt{100 - 100}}{2 \cdot 1}$

Subtract $100$ from $100$ .

$r = \frac{10 \pm \sqrt{0}}{2 \cdot 1}$

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

$r = \frac{10 \pm \sqrt{0^{2}}}{2 \cdot 1}$

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

$r = \frac{10 \pm 0}{2 \cdot 1}$

$10$ plus or minus $0$ is $10$ .

$r = \frac{10}{2 \cdot 1}$

Multiply $2$ by $1$ .

$r = \frac{10}{2}$

Divide $10$ by $2$ .

$r = 5$

Step 4

The final answer is the combination of both solutions.

$r = 5$ Double roots