Solve Using the Quadratic Formula x(x-3)=4

$x (x - 3) = 4$

Step 1

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

Simplify the left side.

Simplify $x (x - 3)$ .

Apply the distributive property.

$x \cdot x + x \cdot - 3 = 4$

Simplify the expression.

Multiply $x$ by $x$ .

$x^{2} + x \cdot - 3 = 4$

Move $- 3$ to the left of $x$ .

$x^{2} - 3 x = 4$

Subtract $4$ from both sides of the equation.

$x^{2} - 3 x - 4 = 0$

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 $a = 1$ , $b = - 3$ , and $c = - 4$ into the quadratic formula and solve for $x$ .

$\frac{3 \pm \sqrt{{(- 3)}^{2} - 4 \cdot (1 \cdot - 4)}}{2 \cdot 1}$

Step 4

Simplify.

Simplify the numerator.

Raise $- 3$ to the power of $2$ .

$x = \frac{3 \pm \sqrt{9 - 4 \cdot 1 \cdot - 4}}{2 \cdot 1}$

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

Multiply $- 4$ by $1$ .

$x = \frac{3 \pm \sqrt{9 - 4 \cdot - 4}}{2 \cdot 1}$

Multiply $- 4$ by $- 4$ .

$x = \frac{3 \pm \sqrt{9 + 16}}{2 \cdot 1}$

Add $9$ and $16$ .

$x = \frac{3 \pm \sqrt{25}}{2 \cdot 1}$

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

$x = \frac{3 \pm \sqrt{5^{2}}}{2 \cdot 1}$

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

$x = \frac{3 \pm 5}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{3 \pm 5}{2}$

Step 5

The final answer is the combination of both solutions.

$x = 4, - 1$