Question 4356
{{{ (x-3) ^2 = 36}}}


Take the square root of both sides to "undo" the square.


{{{sqrt ((x-3)^2) = 0 +- sqrt(36)}}}  
{{{ x - 3 = 0 +- 6}}}


To solve for x, add +3 to each side of the equation:
{{{ x -3 + 3 = 3 +- 6}}}
{{{x = 3 +- 6}}}

In the first case, this means x = 3 + 6, which is x = 9.
In the second case, this means x= 3 - 6, which is x = -3.


Check your answers in the original equation:
{{{ (x-3) ^2 = 36}}}

For x = 9,  {{{ (9-3) ^2 = 36}}}

For x = -3, {{{ (-3-3) ^2 = 36}}}


R^2 at SCC