How do you solve by completing the square: #x^2 + 3x - 18 = 0#?

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Answer 1 · 2015-03-30T14:13:23

In this way:

#x^2+3x-18=0rArrx^2+3x+(3/2)^2-(3/2)^2-18=0rArr#

#(x^2+3x+9/4)-9/4-18=0rArr(x+3/2)^2=18+9/4rArr#

#(x+3/2)^2=(72+9)/4rArr(x+3/2)^2=81/4rArr#

#x+3/2=+-9/2rArrx=-3/2+-9/2rArr#

#x_1=-3/2-9/2=-12/2=-6#

#x_2=-3/2+9/2=6/2=3#.