Solve Using the Quadratic Formula 3x^2-4x+1=0

$3 x^{2} - 4 x + 1 = 0$

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 $a = 3$ , $b = - 4$ , and $c = 1$ into the quadratic formula and solve for $x$ .

$\frac{4 \pm \sqrt{{(- 4)}^{2} - 4 \cdot (3 \cdot 1)}}{2 \cdot 3}$

Step 3

Simplify.

Simplify the numerator.

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

$x = \frac{4 \pm \sqrt{16 - 4 \cdot 3 \cdot 1}}{2 \cdot 3}$

Multiply $- 4 \cdot 3 \cdot 1$ .

Multiply $- 4$ by $3$ .

$x = \frac{4 \pm \sqrt{16 - 12 \cdot 1}}{2 \cdot 3}$

Multiply $- 12$ by $1$ .

$x = \frac{4 \pm \sqrt{16 - 12}}{2 \cdot 3}$

Subtract $12$ from $16$ .

$x = \frac{4 \pm \sqrt{4}}{2 \cdot 3}$

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

$x = \frac{4 \pm \sqrt{2^{2}}}{2 \cdot 3}$

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

$x = \frac{4 \pm 2}{2 \cdot 3}$

Multiply $2$ by $3$ .

$x = \frac{4 \pm 2}{6}$

Simplify $\frac{4 \pm 2}{6}$ .

$x = \frac{2 \pm 1}{3}$

Step 4

The final answer is the combination of both solutions.

$x = 1, \frac{1}{3}$