Solve Using the Quadratic Formula k^2-4k+1=-5

$k^{2} - 4 k + 1 = - 5$

Step 1

Move all terms to the left side of the equation and simplify.

Add $5$ to both sides of the equation.

$k^{2} - 4 k + 1 + 5 = 0$

Add $1$ and $5$ .

$k^{2} - 4 k + 6 = 0$

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 $a = 1$ , $b = - 4$ , and $c = 6$ into the quadratic formula and solve for $k$ .

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

Step 4

Simplify.

Simplify the numerator.

Raise $- 4$ to the power of $2$ .

$k = \frac{4 \pm \sqrt{16 - 4 \cdot 1 \cdot 6}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 6$ .

Multiply $- 4$ by $1$ .

$k = \frac{4 \pm \sqrt{16 - 4 \cdot 6}}{2 \cdot 1}$

Multiply $- 4$ by $6$ .

$k = \frac{4 \pm \sqrt{16 - 24}}{2 \cdot 1}$

Subtract $24$ from $16$ .

$k = \frac{4 \pm \sqrt{- 8}}{2 \cdot 1}$

Rewrite $- 8$ as $- 1 (8)$ .

$k = \frac{4 \pm \sqrt{- 1 \cdot 8}}{2 \cdot 1}$

Rewrite $\sqrt{- 1 (8)}$ as $\sqrt{- 1} \cdot \sqrt{8}$ .

$k = \frac{4 \pm \sqrt{- 1} \cdot \sqrt{8}}{2 \cdot 1}$

Rewrite $\sqrt{- 1}$ as $i$ .

$k = \frac{4 \pm i \cdot \sqrt{8}}{2 \cdot 1}$

Rewrite $8$ as $2^{2} \cdot 2$ .

Factor $4$ out of $8$ .

$k = \frac{4 \pm i \cdot \sqrt{4 (2)}}{2 \cdot 1}$

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

$k = \frac{4 \pm i \cdot \sqrt{2^{2} \cdot 2}}{2 \cdot 1}$

Pull terms out from under the radical.

$k = \frac{4 \pm i \cdot (2 \sqrt{2})}{2 \cdot 1}$

Move $2$ to the left of $i$ .

$k = \frac{4 \pm 2 i \sqrt{2}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$k = \frac{4 \pm 2 i \sqrt{2}}{2}$

Simplify $\frac{4 \pm 2 i \sqrt{2}}{2}$ .

$k = 2 \pm i \sqrt{2}$