Solve Using the Quadratic Formula 4x^2-5x+1=0

$4 x^{2} - 5 x + 1 = 0$

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 $a = 4$ , $b = - 5$ , and $c = 1$ into the quadratic formula and solve for $x$ .

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

Step 3

Simplify.

Simplify the numerator.

Raise $- 5$ to the power of $2$ .

$x = \frac{5 \pm \sqrt{25 - 4 \cdot 4 \cdot 1}}{2 \cdot 4}$

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

Multiply $- 4$ by $4$ .

$x = \frac{5 \pm \sqrt{25 - 16 \cdot 1}}{2 \cdot 4}$

Multiply $- 16$ by $1$ .

$x = \frac{5 \pm \sqrt{25 - 16}}{2 \cdot 4}$

Subtract $16$ from $25$ .

$x = \frac{5 \pm \sqrt{9}}{2 \cdot 4}$

Rewrite $9$ as $3^{2}$ .

$x = \frac{5 \pm \sqrt{3^{2}}}{2 \cdot 4}$

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

$x = \frac{5 \pm 3}{2 \cdot 4}$

Multiply $2$ by $4$ .

$x = \frac{5 \pm 3}{8}$

Step 4

The final answer is the combination of both solutions.

$x = 1, \frac{1}{4}$