How do you solve #x^2 + 3x - 5 = 0# by completing the square?

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Answer 1 · 2017-04-06T16:06:48

#x=-3/2+-sqrt29/4#

Add #5# on both sides:

#x^2+3x=5#

Now add #(b/2a)^2# on both sides where #a=1# and #b=3#

#x^2+3x+9/4=5+9/4#

Rewrite:

#(x+3/2)^2=29/4#

Square root both sides. Don't forget that it's #+-# the square root:

#x+3/2=+-sqrt(29/4)#

Subtract #3/2# on both sides:

#x=-3/2+-sqrt(29/4)#