How do you solve #x^2 + 3x - (7/4) = 0# by completing the square?

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Answer 1 · 2018-04-12T15:17:12

#x = -1.5+-2#

#x^2+3x−7/4=0#

#(x+3/2)^2 - (3/2)^2 - 7/4 = 0#

#(x+3/2)^2 - 9/4 - 7/4 = 0#

#(x+3/2)^2 - 16/4 = 0#

#(x+3/2)^2 = 4#

#x+3/2 = sqrt4#

#color(red)(x= -3/2+-2)#

#color(red)(x=0.5 or x = -3.5)#