Solve Using the Quadratic Formula 5r^2=26r+24

$5 r^{2} = 26 r + 24$

Step 1

Move all the expressions to the left side of the equation.

Subtract $26 r$ from both sides of the equation.

$5 r^{2} - 26 r = 24$

Subtract $24$ from both sides of the equation.

$5 r^{2} - 26 r - 24 = 0$

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 $a = 5$ , $b = - 26$ , and $c = - 24$ into the quadratic formula and solve for $r$ .

$\frac{26 \pm \sqrt{{(- 26)}^{2} - 4 \cdot (5 \cdot - 24)}}{2 \cdot 5}$

Step 4

Simplify.

Simplify the numerator.

Raise $- 26$ to the power of $2$ .

$r = \frac{26 \pm \sqrt{676 - 4 \cdot 5 \cdot - 24}}{2 \cdot 5}$

Multiply $- 4 \cdot 5 \cdot - 24$ .

Multiply $- 4$ by $5$ .

$r = \frac{26 \pm \sqrt{676 - 20 \cdot - 24}}{2 \cdot 5}$

Multiply $- 20$ by $- 24$ .

$r = \frac{26 \pm \sqrt{676 + 480}}{2 \cdot 5}$

Add $676$ and $480$ .

$r = \frac{26 \pm \sqrt{1156}}{2 \cdot 5}$

Rewrite $1156$ as $34^{2}$ .

$r = \frac{26 \pm \sqrt{34^{2}}}{2 \cdot 5}$

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

$r = \frac{26 \pm 34}{2 \cdot 5}$

Multiply $2$ by $5$ .

$r = \frac{26 \pm 34}{10}$

Simplify $\frac{26 \pm 34}{10}$ .

$r = \frac{13 \pm 17}{5}$

Step 5

The final answer is the combination of both solutions.

$r = 6, - \frac{4}{5}$