Solve Using the Quadratic Formula 9x^2-3x-7=0

$9 x^{2} - 3 x - 7 = 0$

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 $a = 9$ , $b = - 3$ , and $c = - 7$ into the quadratic formula and solve for $x$ .

$\frac{3 \pm \sqrt{{(- 3)}^{2} - 4 \cdot (9 \cdot - 7)}}{2 \cdot 9}$

Step 3

Simplify.

Simplify the numerator.

Raise $- 3$ to the power of $2$ .

$x = \frac{3 \pm \sqrt{9 - 4 \cdot 9 \cdot - 7}}{2 \cdot 9}$

Multiply $- 4 \cdot 9 \cdot - 7$ .

Multiply $- 4$ by $9$ .

$x = \frac{3 \pm \sqrt{9 - 36 \cdot - 7}}{2 \cdot 9}$

Multiply $- 36$ by $- 7$ .

$x = \frac{3 \pm \sqrt{9 + 252}}{2 \cdot 9}$

Add $9$ and $252$ .

$x = \frac{3 \pm \sqrt{261}}{2 \cdot 9}$

Rewrite $261$ as $3^{2} \cdot 29$ .

Factor $9$ out of $261$ .

$x = \frac{3 \pm \sqrt{9 (29)}}{2 \cdot 9}$

Rewrite $9$ as $3^{2}$ .

$x = \frac{3 \pm \sqrt{3^{2} \cdot 29}}{2 \cdot 9}$

Pull terms out from under the radical.

$x = \frac{3 \pm 3 \sqrt{29}}{2 \cdot 9}$

Multiply $2$ by $9$ .

$x = \frac{3 \pm 3 \sqrt{29}}{18}$

Simplify $\frac{3 \pm 3 \sqrt{29}}{18}$ .

$x = \frac{1 \pm \sqrt{29}}{6}$

Step 4

The result can be shown in multiple forms.

Exact Form:

$x = \frac{1 \pm \sqrt{29}}{6}$

Decimal Form:

$x = 1.06419413 \dots, - 0.73086080 \dots$