Solve Using the Quadratic Formula 5x^2-6x+1=0

$5 x^{2} - 6 x + 1 = 0$

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 $a = 5$ , $b = - 6$ , and $c = 1$ into the quadratic formula and solve for $x$ .

$\frac{6 \pm \sqrt{{(- 6)}^{2} - 4 \cdot (5 \cdot 1)}}{2 \cdot 5}$

Step 3

Simplify.

Simplify the numerator.

Raise $- 6$ to the power of $2$ .

$x = \frac{6 \pm \sqrt{36 - 4 \cdot 5 \cdot 1}}{2 \cdot 5}$

Multiply $- 4 \cdot 5 \cdot 1$ .

Multiply $- 4$ by $5$ .

$x = \frac{6 \pm \sqrt{36 - 20 \cdot 1}}{2 \cdot 5}$

Multiply $- 20$ by $1$ .

$x = \frac{6 \pm \sqrt{36 - 20}}{2 \cdot 5}$

Subtract $20$ from $36$ .

$x = \frac{6 \pm \sqrt{16}}{2 \cdot 5}$

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

$x = \frac{6 \pm \sqrt{4^{2}}}{2 \cdot 5}$

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

$x = \frac{6 \pm 4}{2 \cdot 5}$

Multiply $2$ by $5$ .

$x = \frac{6 \pm 4}{10}$

Simplify $\frac{6 \pm 4}{10}$ .

$x = \frac{3 \pm 2}{5}$

Step 4

The final answer is the combination of both solutions.

$x = 1, \frac{1}{5}$