Solve Using the Quadratic Formula x(x-4)=5

$x (x - 4) = 5$

Step 1

Move all terms to the left side of the equation and simplify.

Simplify the left side.

Simplify $x (x - 4)$ .

Apply the distributive property.

$x \cdot x + x \cdot - 4 = 5$

Simplify the expression.

Multiply $x$ by $x$ .

$x^{2} + x \cdot - 4 = 5$

Move $- 4$ to the left of $x$ .

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

Subtract $5$ from both sides of the equation.

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

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

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

Step 4

Simplify.

Simplify the numerator.

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

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

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

Multiply $- 4$ by $1$ .

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

Multiply $- 4$ by $- 5$ .

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

Add $16$ and $20$ .

$x = \frac{4 \pm \sqrt{36}}{2 \cdot 1}$

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

$x = \frac{4 \pm \sqrt{6^{2}}}{2 \cdot 1}$

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

$x = \frac{4 \pm 6}{2 \cdot 1}$

Multiply $2$ by $1$ .

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

Simplify $\frac{4 \pm 6}{2}$ .

$x = 2 \pm 3$

Step 5

The final answer is the combination of both solutions.

$x = 5, - 1$