Solve Using the Quadratic Formula 25x^2-20x+4=0

$25 x^{2} - 20 x + 4 = 0$

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 $a = 25$ , $b = - 20$ , and $c = 4$ into the quadratic formula and solve for $x$ .

$\frac{20 \pm \sqrt{{(- 20)}^{2} - 4 \cdot (25 \cdot 4)}}{2 \cdot 25}$

Step 3

Simplify.

Simplify the numerator.

Raise $- 20$ to the power of $2$ .

$x = \frac{20 \pm \sqrt{400 - 4 \cdot 25 \cdot 4}}{2 \cdot 25}$

Multiply $- 4 \cdot 25 \cdot 4$ .

Multiply $- 4$ by $25$ .

$x = \frac{20 \pm \sqrt{400 - 100 \cdot 4}}{2 \cdot 25}$

Multiply $- 100$ by $4$ .

$x = \frac{20 \pm \sqrt{400 - 400}}{2 \cdot 25}$

Subtract $400$ from $400$ .

$x = \frac{20 \pm \sqrt{0}}{2 \cdot 25}$

Rewrite $0$ as $0^{2}$ .

$x = \frac{20 \pm \sqrt{0^{2}}}{2 \cdot 25}$

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

$x = \frac{20 \pm 0}{2 \cdot 25}$

$20$ plus or minus $0$ is $20$ .

$x = \frac{20}{2 \cdot 25}$

Multiply $2$ by $25$ .

$x = \frac{20}{50}$

Cancel the common factor of $20$ and $50$ .

Factor $10$ out of $20$ .

$x = \frac{10 (2)}{50}$

Cancel the common factors.

Factor $10$ out of $50$ .

$x = \frac{10 \cdot 2}{10 \cdot 5}$

Cancel the common factor.

$x = \frac{10 \cdot 2}{10 \cdot 5}$

Rewrite the expression.

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

Step 4

The final answer is the combination of both solutions.

$x = \frac{2}{5}$ Double roots