Find All Complex Solutions 40x^2+1=12x

$40 x^{2} + 1 = 12 x$

Step 1

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

$40 x^{2} + 1 - 12 x = 0$

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 $a = 40$ , $b = - 12$ , and $c = 1$ into the quadratic formula and solve for $x$ .

$\frac{12 \pm \sqrt{{(- 12)}^{2} - 4 \cdot (40 \cdot 1)}}{2 \cdot 40}$

Step 4

Simplify.

Simplify the numerator.

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

$x = \frac{12 \pm \sqrt{144 - 4 \cdot 40 \cdot 1}}{2 \cdot 40}$

Multiply $- 4 \cdot 40 \cdot 1$ .

Multiply $- 4$ by $40$ .

$x = \frac{12 \pm \sqrt{144 - 160 \cdot 1}}{2 \cdot 40}$

Multiply $- 160$ by $1$ .

$x = \frac{12 \pm \sqrt{144 - 160}}{2 \cdot 40}$

Subtract $160$ from $144$ .

$x = \frac{12 \pm \sqrt{- 16}}{2 \cdot 40}$

Rewrite $- 16$ as $- 1 (16)$ .

$x = \frac{12 \pm \sqrt{- 1 \cdot 16}}{2 \cdot 40}$

Rewrite $\sqrt{- 1 (16)}$ as $\sqrt{- 1} \cdot \sqrt{16}$ .

$x = \frac{12 \pm \sqrt{- 1} \cdot \sqrt{16}}{2 \cdot 40}$

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

$x = \frac{12 \pm i \cdot \sqrt{16}}{2 \cdot 40}$

Rewrite $16$ as $4^{2}$ .

$x = \frac{12 \pm i \cdot \sqrt{4^{2}}}{2 \cdot 40}$

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

$x = \frac{12 \pm i \cdot 4}{2 \cdot 40}$

Move $4$ to the left of $i$ .

$x = \frac{12 \pm 4 i}{2 \cdot 40}$

Multiply $2$ by $40$ .

$x = \frac{12 \pm 4 i}{80}$

Simplify $\frac{12 \pm 4 i}{80}$ .

$x = \frac{3 \pm i}{20}$