Solve Using the Quadratic Formula 25x^2-20x+3=0

$25 x^{2} - 20 x + 3 = 0$

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 $c = 3$ into the quadratic formula and solve for $x$ .

$\frac{20 \pm \sqrt{{(- 20)}^{2} - 4 \cdot (25 \cdot 3)}}{2 \cdot 25}$

Step 3

Simplify.

Simplify the numerator.

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

$x = \frac{20 \pm \sqrt{400 - 4 \cdot 25 \cdot 3}}{2 \cdot 25}$

Multiply $- 4 \cdot 25 \cdot 3$ .

Multiply $- 4$ by $25$ .

$x = \frac{20 \pm \sqrt{400 - 100 \cdot 3}}{2 \cdot 25}$

Multiply $- 100$ by $3$ .

$x = \frac{20 \pm \sqrt{400 - 300}}{2 \cdot 25}$

Subtract $300$ from $400$ .

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

Rewrite $100$ as $10^{2}$ .

$x = \frac{20 \pm \sqrt{10^{2}}}{2 \cdot 25}$

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

$x = \frac{20 \pm 10}{2 \cdot 25}$

Multiply $2$ by $25$ .

$x = \frac{20 \pm 10}{50}$

Simplify $\frac{20 \pm 10}{50}$ .

$x = \frac{2 \pm 1}{5}$

Step 4

The final answer is the combination of both solutions.

$x = \frac{3}{5}, \frac{1}{5}$