Solve Using the Quadratic Formula x^2-8x+41=0

$x^{2} - 8 x + 41 = 0$

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 $a = 1$ , $b = - 8$ , and $c = 41$ into the quadratic formula and solve for $x$ .

$\frac{8 \pm \sqrt{{(- 8)}^{2} - 4 \cdot (1 \cdot 41)}}{2 \cdot 1}$

Step 3

Simplify.

Simplify the numerator.

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

$x = \frac{8 \pm \sqrt{64 - 4 \cdot 1 \cdot 41}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 41$ .

Multiply $- 4$ by $1$ .

$x = \frac{8 \pm \sqrt{64 - 4 \cdot 41}}{2 \cdot 1}$

Multiply $- 4$ by $41$ .

$x = \frac{8 \pm \sqrt{64 - 164}}{2 \cdot 1}$

Subtract $164$ from $64$ .

$x = \frac{8 \pm \sqrt{- 100}}{2 \cdot 1}$

Rewrite $- 100$ as $- 1 (100)$ .

$x = \frac{8 \pm \sqrt{- 1 \cdot 100}}{2 \cdot 1}$

Rewrite $\sqrt{- 1 (100)}$ as $\sqrt{- 1} \cdot \sqrt{100}$ .

$x = \frac{8 \pm \sqrt{- 1} \cdot \sqrt{100}}{2 \cdot 1}$

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

$x = \frac{8 \pm i \cdot \sqrt{100}}{2 \cdot 1}$

Rewrite $100$ as $10^{2}$ .

$x = \frac{8 \pm i \cdot \sqrt{10^{2}}}{2 \cdot 1}$

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

$x = \frac{8 \pm i \cdot 10}{2 \cdot 1}$

Move $10$ to the left of $i$ .

$x = \frac{8 \pm 10 i}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{8 \pm 10 i}{2}$

Simplify $\frac{8 \pm 10 i}{2}$ .

$x = 4 \pm 5 i$