Find All Complex Solutions (x+3)^2-4=0

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

Step 1

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 2

Substitute the values $a = 1$ , $b = 6$ , and $c = 5$ into the quadratic formula and solve for $x$ .

$\frac{- 6 \pm \sqrt{6^{2} - 4 \cdot (1 \cdot 5)}}{2 \cdot 1}$

Step 3

Simplify.

Simplify the numerator.

Raise $6$ to the power of $2$ .

$x = \frac{- 6 \pm \sqrt{36 - 4 \cdot 1 \cdot 5}}{2 \cdot 1}$

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

Multiply $- 4$ by $1$ .

$x = \frac{- 6 \pm \sqrt{36 - 4 \cdot 5}}{2 \cdot 1}$

Multiply $- 4$ by $5$ .

$x = \frac{- 6 \pm \sqrt{36 - 20}}{2 \cdot 1}$

Subtract $20$ from $36$ .

$x = \frac{- 6 \pm \sqrt{16}}{2 \cdot 1}$

Rewrite $16$ as $4^{2}$ .

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

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

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

Multiply $2$ by $1$ .

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

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

$x = - 3 \pm 2$

Step 4

The final answer is the combination of both solutions.

$x = - 1, - 5$