Solve Using the Quadratic Formula x(x-5)=0

$x (x - 5) = 0$

Step 1

Simplify the left side.

Simplify $x (x - 5)$ .

Apply the distributive property.

$x \cdot x + x \cdot - 5 = 0$

Simplify the expression.

Multiply $x$ by $x$ .

$x^{2} + x \cdot - 5 = 0$

Move $- 5$ to the left of $x$ .

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

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 $a = 1$ , $b = - 5$ , and $c = 0$ into the quadratic formula and solve for $x$ .

$\frac{5 \pm \sqrt{{(- 5)}^{2} - 4 \cdot (1 \cdot 0)}}{2 \cdot 1}$

Step 4

Simplify.

Simplify the numerator.

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

$x = \frac{5 \pm \sqrt{25 - 4 \cdot 1 \cdot 0}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 0$ .

Multiply $- 4$ by $1$ .

$x = \frac{5 \pm \sqrt{25 - 4 \cdot 0}}{2 \cdot 1}$

Multiply $- 4$ by $0$ .

$x = \frac{5 \pm \sqrt{25 + 0}}{2 \cdot 1}$

Add $25$ and $0$ .

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

Rewrite $25$ as $5^{2}$ .

$x = \frac{5 \pm \sqrt{5^{2}}}{2 \cdot 1}$

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

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

Multiply $2$ by $1$ .

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

Step 5

The final answer is the combination of both solutions.

$x = 5, 0$