Solve Using the Quadratic Formula r^2-30r+29=0

$r^{2} - 30 r + 29 = 0$

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 $a = 1$ , $b = - 30$ , and $c = 29$ into the quadratic formula and solve for $r$ .

$\frac{30 \pm \sqrt{{(- 30)}^{2} - 4 \cdot (1 \cdot 29)}}{2 \cdot 1}$

Step 3

Simplify.

Simplify the numerator.

Raise $- 30$ to the power of $2$ .

$r = \frac{30 \pm \sqrt{900 - 4 \cdot 1 \cdot 29}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 29$ .

Multiply $- 4$ by $1$ .

$r = \frac{30 \pm \sqrt{900 - 4 \cdot 29}}{2 \cdot 1}$

Multiply $- 4$ by $29$ .

$r = \frac{30 \pm \sqrt{900 - 116}}{2 \cdot 1}$

Subtract $116$ from $900$ .

$r = \frac{30 \pm \sqrt{784}}{2 \cdot 1}$

Rewrite $784$ as $28^{2}$ .

$r = \frac{30 \pm \sqrt{28^{2}}}{2 \cdot 1}$

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

$r = \frac{30 \pm 28}{2 \cdot 1}$

Multiply $2$ by $1$ .

$r = \frac{30 \pm 28}{2}$

Simplify $\frac{30 \pm 28}{2}$ .

$r = 15 \pm 14$

Step 4

The final answer is the combination of both solutions.

$r = 29, 1$