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

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

Step 1

Subtract $1$ from both sides of the equation.

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

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

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

Step 4

Simplify.

Simplify the numerator.

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

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

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

Multiply $- 4$ by $5$ .

$x = \frac{4 \pm \sqrt{16 - 20 \cdot - 1}}{2 \cdot 5}$

Multiply $- 20$ by $- 1$ .

$x = \frac{4 \pm \sqrt{16 + 20}}{2 \cdot 5}$

Add $16$ and $20$ .

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

Rewrite $36$ as $6^{2}$ .

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

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

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

Multiply $2$ by $5$ .

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

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

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

Step 5

The final answer is the combination of both solutions.

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