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

$6 a^{2} - a - 5 = 0$

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 $a = 6$ , $b = - 1$ , and $c = - 5$ into the quadratic formula and solve for $a$ .

$\frac{1 \pm \sqrt{{(- 1)}^{2} - 4 \cdot (6 \cdot - 5)}}{2 \cdot 6}$

Step 3

Simplify.

Simplify the numerator.

Raise $- 1$ to the power of $2$ .

$a = \frac{1 \pm \sqrt{1 - 4 \cdot 6 \cdot - 5}}{2 \cdot 6}$

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

Multiply $- 4$ by $6$ .

$a = \frac{1 \pm \sqrt{1 - 24 \cdot - 5}}{2 \cdot 6}$

Multiply $- 24$ by $- 5$ .

$a = \frac{1 \pm \sqrt{1 + 120}}{2 \cdot 6}$

Add $1$ and $120$ .

$a = \frac{1 \pm \sqrt{121}}{2 \cdot 6}$

Rewrite $121$ as $11^{2}$ .

$a = \frac{1 \pm \sqrt{11^{2}}}{2 \cdot 6}$

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

$a = \frac{1 \pm 11}{2 \cdot 6}$

Multiply $2$ by $6$ .

$a = \frac{1 \pm 11}{12}$

Step 4

The final answer is the combination of both solutions.

$a = 1, - \frac{5}{6}$