Solve for z z^2+6z=1

$z^{2} + 6 z = 1$

Step 1

Subtract $1$ from both sides of the equation.

$z^{2} + 6 z - 1 = 0$

Step 2

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 3

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

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

Step 4

Simplify.

Simplify the numerator.

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

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

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

Multiply $- 4$ by $1$ .

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

Multiply $- 4$ by $- 1$ .

$z = \frac{- 6 \pm \sqrt{36 + 4}}{2 \cdot 1}$

Add $36$ and $4$ .

$z = \frac{- 6 \pm \sqrt{40}}{2 \cdot 1}$

Rewrite $40$ as $2^{2} \cdot 10$ .

Factor $4$ out of $40$ .

$z = \frac{- 6 \pm \sqrt{4 (10)}}{2 \cdot 1}$

Rewrite $4$ as $2^{2}$ .

$z = \frac{- 6 \pm \sqrt{2^{2} \cdot 10}}{2 \cdot 1}$

Pull terms out from under the radical.

$z = \frac{- 6 \pm 2 \sqrt{10}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$z = \frac{- 6 \pm 2 \sqrt{10}}{2}$

Simplify $\frac{- 6 \pm 2 \sqrt{10}}{2}$ .

$z = - 3 \pm \sqrt{10}$

Step 5

The final answer is the combination of both solutions.

$z = - 3 + \sqrt{10}, - 3 - \sqrt{10}$

Step 6

The result can be shown in multiple forms.

Exact Form:

$z = - 3 + \sqrt{10}, - 3 - \sqrt{10}$

Decimal Form:

$z = 0.16227766 \dots, - 6.16227766 \dots$