Find the Roots (Zeros) 12x^2=4x-1

$12 x^{2} = 4 x - 1$

Subtract $4 x$ from both sides of the equation.

$12 x^{2} - 4 x = - 1$

Add $1$ to both sides of the equation.

$12 x^{2} - 4 x + 1 = 0$

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Substitute the values $a = 12$ , $b = - 4$ , and $c = 1$ into the quadratic formula and solve for $x$ .

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

Simplify.

Simplify the numerator.

Raise $- 4$ to the power of $2$ .

$x = \frac{4 \pm \sqrt{16 - 4 \cdot 12 \cdot 1}}{2 \cdot 12}$

Multiply $- 4$ by $12$ .

$x = \frac{4 \pm \sqrt{16 - 48 \cdot 1}}{2 \cdot 12}$

Multiply $- 48$ by $1$ .

$x = \frac{4 \pm \sqrt{16 - 48}}{2 \cdot 12}$

Subtract $48$ from $16$ .

$x = \frac{4 \pm \sqrt{- 32}}{2 \cdot 12}$

Rewrite $- 32$ as $- 1 (32)$ .

$x = \frac{4 \pm \sqrt{- 1 \cdot 32}}{2 \cdot 12}$

Rewrite $\sqrt{- 1 (32)}$ as $\sqrt{- 1} \cdot \sqrt{32}$ .

$x = \frac{4 \pm \sqrt{- 1} \cdot \sqrt{32}}{2 \cdot 12}$

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

$x = \frac{4 \pm i \cdot \sqrt{32}}{2 \cdot 12}$

Rewrite $32$ as $4^{2} \cdot 2$ .

Factor $16$ out of $32$ .

$x = \frac{4 \pm i \cdot \sqrt{16 (2)}}{2 \cdot 12}$

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

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

Pull terms out from under the radical.

$x = \frac{4 \pm i \cdot (4 \sqrt{2})}{2 \cdot 12}$

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

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

Multiply $2$ by $12$ .

$x = \frac{4 \pm 4 i \sqrt{2}}{24}$

Simplify $\frac{4 \pm 4 i \sqrt{2}}{24}$ .

$x = \frac{1 \pm i \sqrt{2}}{6}$