There are more discussions, explanations and examples here of Completing the Square
Here is a solution for this question:
#x^2+3x-2=0#
#(x^2+3x color(white)("sssss"))-2=0#
#!/2# of #3# is #3/2#. Square that to get #9/4# add #9/4# to complete the square and subtract to keep the equation balanced:
#(x^2+3x +9/4-9/4)-2=0# Regoup to keep just the perfect square in the parentheses:
#(x^2+3x +9/4)-9/4-2=0#
Factor the perfect square (use the #3/2# from before and "#-#" like in the #+3x#. Also simplify #-9/2-2#
#(x +3/2)^2-9/4-8/4= (x +3/2)^2-17/4= 0#
Solve #(x +3/2)^2-17/4= 0# by "the Square Root Method"
#(x +3/2)^2-17/4= 0#
#(x +3/2)^2= 17/4#
#x +3/2 = +-sqrt(17/4) = +-sqrt17/sqrt4#
#x +3/2 = +- sqrt(17)/2#
#x = -3/2 +- sqrt(17)/2 = (-3 +- sqrt(17))/2 #
(Use whichever form your teacher prefers -- one fraction or two.)
