How do you solve the equation #x^2-3x-7=0# by completing the square?

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Answer 1 · 2018-03-03T08:01:50

#color(blue)(x = 4.54, -1.54#

#x^2 - 3x - 7 = 0#

#x^2 - (2*(3/2)*x) = 7#

Adding #(3/2)^2# to both sides,

#x^2 - (2 * 3/2) * x) + (3/2)^2 = 7 + (3/2)^2 = 37/4#

#(x - (3/2))^2 = (sqrt(37/4))^2#

#x =+- sqrt(37/4)+ (3/2)#

#color(blue)(x= 4.54, -1.54#