How do you solve #2x^2+6x-3=0# by completing the square?

Question

14412 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-04-29T16:26:34

In this way:

#2x^2+6x-3=0rArr2(x^2+3x)-3=0rArr#

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

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

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

#2(x+3/2)^2=(9+6)/2rArr(x+3/2)^2=15/4rArr#

#x+3/2=+-sqrt15/2rArrx=-3/2+-sqrt15/2#.