How do you solve #-2x^2+10x=-14# by completing the square?

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Answer 1 · 2018-01-22T08:24:35

#−2x^2+10x=−14#

#2x^2-10x-14=0#

dividing the entire equation by 2,
#x^2-5x-7=0#

now adding and subtract "half of the #x# term squared",
which in this case is, #(5/2)^2#

#(x^2-5x+ (5/2)^2) -7-(5/2)^2=0#

#(x- 5/2)^2 -7-(25/4)=0#

#(x- 5/2)^2 -28/4-25/4=0#

#(x- 5/2)^2 -53/4=0#

#(x- 5/2)^2 =53/4#

#(x- 5/2) =+-sqrt(53/4)#

#(x- 5/2) =+-sqrt53/2#

#x=+-sqrt53/2+ 5/2 #

#x=(5+-sqrt53)/2 #