Question 242158
Take the square root of each side of the equation, giving you


{{{x^2=180}}}
{{{x=0+-sqrt(180)}}}


NOTE:  You really don't need the 0 in front of the +/-, but the algebra.com notation requires that I put it there!


{{{x=0+-sqrt(36)*sqrt(5)}}}
{{{x=6sqrt(5)}}} or {{{x=-6sqrt(5)}}}


Dr. Robert J. Rapalje, Retired
Seminole State College of Florida