Solve Using the Quadratic Formula 7x^2-x-6=0

$7 x^{2} - x - 6 = 0$

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 $a = 7$ , $b = - 1$ , and $c = - 6$ into the quadratic formula and solve for $x$ .

$\frac{1 \pm \sqrt{{(- 1)}^{2} - 4 \cdot (7 \cdot - 6)}}{2 \cdot 7}$

Step 3

Simplify.

Simplify the numerator.

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

$x = \frac{1 \pm \sqrt{1 - 4 \cdot 7 \cdot - 6}}{2 \cdot 7}$

Multiply $- 4 \cdot 7 \cdot - 6$ .

Multiply $- 4$ by $7$ .

$x = \frac{1 \pm \sqrt{1 - 28 \cdot - 6}}{2 \cdot 7}$

Multiply $- 28$ by $- 6$ .

$x = \frac{1 \pm \sqrt{1 + 168}}{2 \cdot 7}$

Add $1$ and $168$ .

$x = \frac{1 \pm \sqrt{169}}{2 \cdot 7}$

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

$x = \frac{1 \pm \sqrt{13^{2}}}{2 \cdot 7}$

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

$x = \frac{1 \pm 13}{2 \cdot 7}$

Multiply $2$ by $7$ .

$x = \frac{1 \pm 13}{14}$

Step 4

The final answer is the combination of both solutions.

$x = 1, - \frac{6}{7}$