Solve Using the Quadratic Formula x^2-5x+25/4=0

$x^{2} - 5 x + \frac{25}{4} = 0$

Step 1

Multiply through by the least common denominator $4$ , then simplify.

Apply the distributive property.

$4 x^{2} + 4 (- 5 x) + 4 (\frac{25}{4}) = 0$

Simplify.

Multiply $- 5$ by $4$ .

$4 x^{2} - 20 x + 4 (\frac{25}{4}) = 0$

Cancel the common factor of $4$ .

Cancel the common factor.

$4 x^{2} - 20 x + 4 (\frac{25}{4}) = 0$

Rewrite the expression.

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

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

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

Step 4

Simplify.

Simplify the numerator.

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

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

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

Multiply $- 4$ by $4$ .

$x = \frac{20 \pm \sqrt{400 - 16 \cdot 25}}{2 \cdot 4}$

Multiply $- 16$ by $25$ .

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

Subtract $400$ from $400$ .

$x = \frac{20 \pm \sqrt{0}}{2 \cdot 4}$

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

$x = \frac{20 \pm \sqrt{0^{2}}}{2 \cdot 4}$

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

$x = \frac{20 \pm 0}{2 \cdot 4}$

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

$x = \frac{20}{2 \cdot 4}$

Multiply $2$ by $4$ .

$x = \frac{20}{8}$

Cancel the common factor of $20$ and $8$ .

Factor $4$ out of $20$ .

$x = \frac{4 (5)}{8}$

Cancel the common factors.

Factor $4$ out of $8$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Step 5

The final answer is the combination of both solutions.

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

Solve Using the Quadratic Formula x^2-5x+25/4=0

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