Find All Complex Solutions x^2+4x=2x-7

$x^{2} + 4 x = 2 x - 7$

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 $2 x$ from both sides of the equation.

$x^{2} + 4 x - 2 x = - 7$

Add $7$ to both sides of the equation.

$x^{2} + 4 x - 2 x + 7 = 0$

Subtract $2 x$ from $4 x$ .

$x^{2} + 2 x + 7 = 0$

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 $a = 1$ , $b = 2$ , and $c = 7$ into the quadratic formula and solve for $x$ .

$\frac{- 2 \pm \sqrt{2^{2} - 4 \cdot (1 \cdot 7)}}{2 \cdot 1}$

Step 4

Simplify.

Simplify the numerator.

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

$x = \frac{- 2 \pm \sqrt{4 - 4 \cdot 1 \cdot 7}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 7$ .

Multiply $- 4$ by $1$ .

$x = \frac{- 2 \pm \sqrt{4 - 4 \cdot 7}}{2 \cdot 1}$

Multiply $- 4$ by $7$ .

$x = \frac{- 2 \pm \sqrt{4 - 28}}{2 \cdot 1}$

Subtract $28$ from $4$ .

$x = \frac{- 2 \pm \sqrt{- 24}}{2 \cdot 1}$

Rewrite $- 24$ as $- 1 (24)$ .

$x = \frac{- 2 \pm \sqrt{- 1 \cdot 24}}{2 \cdot 1}$

Rewrite $\sqrt{- 1 (24)}$ as $\sqrt{- 1} \cdot \sqrt{24}$ .

$x = \frac{- 2 \pm \sqrt{- 1} \cdot \sqrt{24}}{2 \cdot 1}$

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{- 2 \pm i \cdot \sqrt{24}}{2 \cdot 1}$

Rewrite $24$ as $2^{2} \cdot 6$ .

Factor $4$ out of $24$ .

$x = \frac{- 2 \pm i \cdot \sqrt{4 (6)}}{2 \cdot 1}$

Rewrite $4$ as $2^{2}$ .

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

Pull terms out from under the radical.

$x = \frac{- 2 \pm i \cdot (2 \sqrt{6})}{2 \cdot 1}$

Move $2$ to the left of $i$ .

$x = \frac{- 2 \pm 2 i \sqrt{6}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 2 \pm 2 i \sqrt{6}}{2}$

Simplify $\frac{- 2 \pm 2 i \sqrt{6}}{2}$ .

$x = - 1 \pm i \sqrt{6}$

Find All Complex Solutions x^2+4x=2x-7

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