Solve Using the Quadratic Formula 6x^2-13x-5=0

$6 x^{2} - 13 x - 5 = 0$

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 $a = 6$ , $b = - 13$ , and $c = - 5$ into the quadratic formula and solve for $x$ .

$\frac{13 \pm \sqrt{{(- 13)}^{2} - 4 \cdot (6 \cdot - 5)}}{2 \cdot 6}$

Step 3

Simplify.

Simplify the numerator.

Raise $- 13$ to the power of $2$ .

$x = \frac{13 \pm \sqrt{169 - 4 \cdot 6 \cdot - 5}}{2 \cdot 6}$

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

Multiply $- 4$ by $6$ .

$x = \frac{13 \pm \sqrt{169 - 24 \cdot - 5}}{2 \cdot 6}$

Multiply $- 24$ by $- 5$ .

$x = \frac{13 \pm \sqrt{169 + 120}}{2 \cdot 6}$

Add $169$ and $120$ .

$x = \frac{13 \pm \sqrt{289}}{2 \cdot 6}$

Rewrite $289$ as $17^{2}$ .

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

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

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

Multiply $2$ by $6$ .

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

Step 4

The final answer is the combination of both solutions.

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