Solve Using the Quadratic Formula 7x^2+8x+1=0

$7 x^{2} + 8 x + 1 = 0$

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 $a = 7$ , $b = 8$ , and $c = 1$ into the quadratic formula and solve for $x$ .

$\frac{- 8 \pm \sqrt{8^{2} - 4 \cdot (7 \cdot 1)}}{2 \cdot 7}$

Step 3

Simplify.

Simplify the numerator.

Raise $8$ to the power of $2$ .

$x = \frac{- 8 \pm \sqrt{64 - 4 \cdot 7 \cdot 1}}{2 \cdot 7}$

Multiply $- 4 \cdot 7 \cdot 1$ .

Multiply $- 4$ by $7$ .

$x = \frac{- 8 \pm \sqrt{64 - 28 \cdot 1}}{2 \cdot 7}$

Multiply $- 28$ by $1$ .

$x = \frac{- 8 \pm \sqrt{64 - 28}}{2 \cdot 7}$

Subtract $28$ from $64$ .

$x = \frac{- 8 \pm \sqrt{36}}{2 \cdot 7}$

Rewrite $36$ as $6^{2}$ .

$x = \frac{- 8 \pm \sqrt{6^{2}}}{2 \cdot 7}$

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

$x = \frac{- 8 \pm 6}{2 \cdot 7}$

Multiply $2$ by $7$ .

$x = \frac{- 8 \pm 6}{14}$

Simplify $\frac{- 8 \pm 6}{14}$ .

$x = \frac{- 4 \pm 3}{7}$

Step 4

The final answer is the combination of both solutions.

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