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

$q^{2} + 6 q - 7 = 0$

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

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

Step 3

Simplify.

Simplify the numerator.

Raise $6$ to the power of $2$ .

$q = \frac{- 6 \pm \sqrt{36 - 4 \cdot 1 \cdot - 7}}{2 \cdot 1}$

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

Multiply $- 4$ by $1$ .

$q = \frac{- 6 \pm \sqrt{36 - 4 \cdot - 7}}{2 \cdot 1}$

Multiply $- 4$ by $- 7$ .

$q = \frac{- 6 \pm \sqrt{36 + 28}}{2 \cdot 1}$

Add $36$ and $28$ .

$q = \frac{- 6 \pm \sqrt{64}}{2 \cdot 1}$

Rewrite $64$ as $8^{2}$ .

$q = \frac{- 6 \pm \sqrt{8^{2}}}{2 \cdot 1}$

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

$q = \frac{- 6 \pm 8}{2 \cdot 1}$

Multiply $2$ by $1$ .

$q = \frac{- 6 \pm 8}{2}$

Simplify $\frac{- 6 \pm 8}{2}$ .

$q = - 3 \pm 4$

Step 4

The final answer is the combination of both solutions.

$q = 1, - 7$