Find All Complex Solutions 13x^2-5x+1=0

$13 x^{2} - 5 x + 1 = 0$

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 $a = 13$ , $b = - 5$ , and $c = 1$ into the quadratic formula and solve for $x$ .

$\frac{5 \pm \sqrt{{(- 5)}^{2} - 4 \cdot (13 \cdot 1)}}{2 \cdot 13}$

Step 3

Simplify.

Simplify the numerator.

Raise $- 5$ to the power of $2$ .

$x = \frac{5 \pm \sqrt{25 - 4 \cdot 13 \cdot 1}}{2 \cdot 13}$

Multiply $- 4 \cdot 13 \cdot 1$ .

Multiply $- 4$ by $13$ .

$x = \frac{5 \pm \sqrt{25 - 52 \cdot 1}}{2 \cdot 13}$

Multiply $- 52$ by $1$ .

$x = \frac{5 \pm \sqrt{25 - 52}}{2 \cdot 13}$

Subtract $52$ from $25$ .

$x = \frac{5 \pm \sqrt{- 27}}{2 \cdot 13}$

Rewrite $- 27$ as $- 1 (27)$ .

$x = \frac{5 \pm \sqrt{- 1 \cdot 27}}{2 \cdot 13}$

Rewrite $\sqrt{- 1 (27)}$ as $\sqrt{- 1} \cdot \sqrt{27}$ .

$x = \frac{5 \pm \sqrt{- 1} \cdot \sqrt{27}}{2 \cdot 13}$

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

$x = \frac{5 \pm i \cdot \sqrt{27}}{2 \cdot 13}$

Rewrite $27$ as $3^{2} \cdot 3$ .

Factor $9$ out of $27$ .

$x = \frac{5 \pm i \cdot \sqrt{9 (3)}}{2 \cdot 13}$

Rewrite $9$ as $3^{2}$ .

$x = \frac{5 \pm i \cdot \sqrt{3^{2} \cdot 3}}{2 \cdot 13}$

Pull terms out from under the radical.

$x = \frac{5 \pm i \cdot (3 \sqrt{3})}{2 \cdot 13}$

Move $3$ to the left of $i$ .

$x = \frac{5 \pm 3 i \sqrt{3}}{2 \cdot 13}$

Multiply $2$ by $13$ .

$x = \frac{5 \pm 3 i \sqrt{3}}{26}$