Find All Complex Solutions -x^2+4x=x-4

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

Step 1

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

Move all the expressions to the left side of the equation.

Subtract $x$ from both sides of the equation.

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

Add $4$ to both sides of the equation.

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

Subtract $x$ from $4 x$ .

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

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 $- 1$ .

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

Simplify $\frac{- 3 \pm 5}{- 2}$ .

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

Step 5

The final answer is the combination of both solutions.

$x = 4, - 1$