Solve Using the Quadratic Formula 6x^2-17x+5=0

$6 x^{2} - 17 x + 5 = 0$

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 $a = 6$ , $b = - 17$ , and $c = 5$ into the quadratic formula and solve for $x$ .

$\frac{17 \pm \sqrt{{(- 17)}^{2} - 4 \cdot (6 \cdot 5)}}{2 \cdot 6}$

Step 3

Simplify.

Simplify the numerator.

Raise $- 17$ to the power of $2$ .

$x = \frac{17 \pm \sqrt{289 - 4 \cdot 6 \cdot 5}}{2 \cdot 6}$

Multiply $- 4 \cdot 6 \cdot 5$ .

Multiply $- 4$ by $6$ .

$x = \frac{17 \pm \sqrt{289 - 24 \cdot 5}}{2 \cdot 6}$

Multiply $- 24$ by $5$ .

$x = \frac{17 \pm \sqrt{289 - 120}}{2 \cdot 6}$

Subtract $120$ from $289$ .

$x = \frac{17 \pm \sqrt{169}}{2 \cdot 6}$

Rewrite $169$ as $13^{2}$ .

$x = \frac{17 \pm \sqrt{13^{2}}}{2 \cdot 6}$

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

$x = \frac{17 \pm 13}{2 \cdot 6}$

Multiply $2$ by $6$ .

$x = \frac{17 \pm 13}{12}$

Step 4

The final answer is the combination of both solutions.

$x = \frac{5}{2}, \frac{1}{3}$