Solve for x (x^2+2x+6)^(3/2)+1=0

${(x^{2} + 2 x + 6)}^{\frac{3}{2}} + 1 = 0$

Subtract $1$ from both sides of the equation.

${(x^{2} + 2 x + 6)}^{\frac{3}{2}} = - 1$

Raise each side of the equation to the power of $\frac{2}{3}$ to eliminate the fractional exponent on the left side.

${({(x^{2} + 2 x + 6)}^{\frac{3}{2}})}^{\frac{2}{3}} = {(- 1)}^{\frac{2}{3}}$

Simplify the exponent.

Simplify the left side.

Simplify ${({(x^{2} + 2 x + 6)}^{\frac{3}{2}})}^{\frac{2}{3}}$ .

Multiply the exponents in ${({(x^{2} + 2 x + 6)}^{\frac{3}{2}})}^{\frac{2}{3}}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

${(x^{2} + 2 x + 6)}^{\frac{3}{2} \cdot \frac{2}{3}} = {(- 1)}^{\frac{2}{3}}$

Cancel the common factor of $3$ .

Cancel the common factor.

${(x^{2} + 2 x + 6)}^{\frac{3}{2} \cdot \frac{2}{3}} = {(- 1)}^{\frac{2}{3}}$

Rewrite the expression.

${(x^{2} + 2 x + 6)}^{\frac{1}{2} \cdot 2} = {(- 1)}^{\frac{2}{3}}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(x^{2} + 2 x + 6)}^{\frac{1}{2} \cdot 2} = {(- 1)}^{\frac{2}{3}}$

Rewrite the expression.

${(x^{2} + 2 x + 6)}^{1} = {(- 1)}^{\frac{2}{3}}$

Simplify.

$x^{2} + 2 x + 6 = {(- 1)}^{\frac{2}{3}}$

Simplify the right side.

Simplify ${(- 1)}^{\frac{2}{3}}$ .

Simplify the expression.

Rewrite $- 1$ as ${(- 1)}^{3}$ .

$x^{2} + 2 x + 6 = {({(- 1)}^{3})}^{\frac{2}{3}}$

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$x^{2} + 2 x + 6 = {(- 1)}^{3 (\frac{2}{3})}$

Cancel the common factor of $3$ .

Cancel the common factor.

$x^{2} + 2 x + 6 = {(- 1)}^{3 (\frac{2}{3})}$

Rewrite the expression.

$x^{2} + 2 x + 6 = {(- 1)}^{2}$

Raise $- 1$ to the power of $2$ .

$x^{2} + 2 x + 6 = 1$

Solve for $x$ .

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

Subtract $1$ from both sides of the equation.

$x^{2} + 2 x + 6 - 1 = 0$

Subtract $1$ from $6$ .

$x^{2} + 2 x + 5 = 0$

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 1$ , $b = 2$ , and $c = 5$ into the quadratic formula and solve for $x$ .

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

Simplify.

Simplify the numerator.

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

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

Multiply $- 4$ by $1$ .

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

Multiply $- 4$ by $5$ .

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

Subtract $20$ from $4$ .

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

Rewrite $- 16$ as $- 1 (16)$ .

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

Rewrite $\sqrt{- 1 (16)}$ as $\sqrt{- 1} \cdot \sqrt{16}$ .

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

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

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

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

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

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

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

Move $4$ to the left of $i$ .

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

Multiply $2$ by $1$ .

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

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

$x = - 1 \pm 2 i$

The final answer is the combination of both solutions.

$x = - 1 + 2 i, - 1 - 2 i$

Solve for x (x^2+2x+6)^(3/2)+1=0

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