Find All Complex Solutions 25x^2+1=8x

$25 x^{2} + 1 = 8 x$

Step 1

Subtract $8 x$ from both sides of the equation.

$25 x^{2} + 1 - 8 x = 0$

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 $a = 25$ , $b = - 8$ , and $c = 1$ into the quadratic formula and solve for $x$ .

$\frac{8 \pm \sqrt{{(- 8)}^{2} - 4 \cdot (25 \cdot 1)}}{2 \cdot 25}$

Step 4

Simplify.

Simplify the numerator.

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

$x = \frac{8 \pm \sqrt{64 - 4 \cdot 25 \cdot 1}}{2 \cdot 25}$

Multiply $- 4 \cdot 25 \cdot 1$ .

Multiply $- 4$ by $25$ .

$x = \frac{8 \pm \sqrt{64 - 100 \cdot 1}}{2 \cdot 25}$

Multiply $- 100$ by $1$ .

$x = \frac{8 \pm \sqrt{64 - 100}}{2 \cdot 25}$

Subtract $100$ from $64$ .

$x = \frac{8 \pm \sqrt{- 36}}{2 \cdot 25}$

Rewrite $- 36$ as $- 1 (36)$ .

$x = \frac{8 \pm \sqrt{- 1 \cdot 36}}{2 \cdot 25}$

Rewrite $\sqrt{- 1 (36)}$ as $\sqrt{- 1} \cdot \sqrt{36}$ .

$x = \frac{8 \pm \sqrt{- 1} \cdot \sqrt{36}}{2 \cdot 25}$

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

$x = \frac{8 \pm i \cdot \sqrt{36}}{2 \cdot 25}$

Rewrite $36$ as $6^{2}$ .

$x = \frac{8 \pm i \cdot \sqrt{6^{2}}}{2 \cdot 25}$

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

$x = \frac{8 \pm i \cdot 6}{2 \cdot 25}$

Move $6$ to the left of $i$ .

$x = \frac{8 \pm 6 i}{2 \cdot 25}$

Multiply $2$ by $25$ .

$x = \frac{8 \pm 6 i}{50}$

Simplify $\frac{8 \pm 6 i}{50}$ .

$x = \frac{4 \pm 3 i}{25}$