Chapter 5 - Polynomials and Polynomial Functions - 5-5 Theorems About Roots of Polynomial Equations - Practice and Problem-Solving Exercises - Page 317: 58

$x = ± 3i$

Factor out a $7$ from both terms to get: $7(x^2 + 9) = 0$ We divide both sides by $7$ to get rid of the $7$. We are left with: $x^2 + 9 = 0$ $x^2 = -9$ Take the square root of both sides: $x = \pm\sqrt{-9}$ $x = \pm\sqrt {(-1)(9)}$ The square root of $-1$ is $i$, and the square root of $9$ is $3$: $x = ± 3i$