Find All Complex Solutions 117x^2+1=12x

$117 x^{2} + 1 = 12 x$

Step 1

Subtract $12 x$ from both sides of the equation.

$117 x^{2} + 1 - 12 x = 0$

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 $a = 117$ , $b = - 12$ , and $c = 1$ into the quadratic formula and solve for $x$ .

$\frac{12 \pm \sqrt{{(- 12)}^{2} - 4 \cdot (117 \cdot 1)}}{2 \cdot 117}$

Step 4

Simplify.

Simplify the numerator.

Raise $- 12$ to the power of $2$ .

$x = \frac{12 \pm \sqrt{144 - 4 \cdot 117 \cdot 1}}{2 \cdot 117}$

Multiply $- 4 \cdot 117 \cdot 1$ .

Multiply $- 4$ by $117$ .

$x = \frac{12 \pm \sqrt{144 - 468 \cdot 1}}{2 \cdot 117}$

Multiply $- 468$ by $1$ .

$x = \frac{12 \pm \sqrt{144 - 468}}{2 \cdot 117}$

Subtract $468$ from $144$ .

$x = \frac{12 \pm \sqrt{- 324}}{2 \cdot 117}$

Rewrite $- 324$ as $- 1 (324)$ .

$x = \frac{12 \pm \sqrt{- 1 \cdot 324}}{2 \cdot 117}$

Rewrite $\sqrt{- 1 (324)}$ as $\sqrt{- 1} \cdot \sqrt{324}$ .

$x = \frac{12 \pm \sqrt{- 1} \cdot \sqrt{324}}{2 \cdot 117}$

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

$x = \frac{12 \pm i \cdot \sqrt{324}}{2 \cdot 117}$

Rewrite $324$ as $18^{2}$ .

$x = \frac{12 \pm i \cdot \sqrt{18^{2}}}{2 \cdot 117}$

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

$x = \frac{12 \pm i \cdot 18}{2 \cdot 117}$

Move $18$ to the left of $i$ .

$x = \frac{12 \pm 18 i}{2 \cdot 117}$

Multiply $2$ by $117$ .

$x = \frac{12 \pm 18 i}{234}$

Simplify $\frac{12 \pm 18 i}{234}$ .

$x = \frac{2 \pm 3 i}{39}$