How do you solve using the completing the square method #2x^2+7x-5=0#?

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Answer 1 · 2018-03-03T11:31:37

#color(blue)(x = (1/4) (3sqrt11 - 7), -(1/4)(3sqrt11 + 7)#

#2x2 + 7x - 5 = 0#

Dividing by 2, #x2 + 2 * (7/4) x = 5/2#

Adding #(7/4)^2# to both sides,

#x^2 = (2*(7/4)x + (7/4)^2 = 5/2 + (7/4)^2 = 99/16#

#(x+7/4)^2 = (sqrt(99/16))^2#

#x + 7/4 = +- sqrt(99/16) #

#x = sqrt(99/16) - 7/4, -sqrt(99/16 ) - 7/4#

#x = (1/4) (3sqrt11 - 7), -(1/4)(3sqrt11 + 7)#