How do you solve #2x^2 - 12x + 10 = 0#?

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Answer 1 · 2015-11-04T17:31:05

The two solutions are #x=1# and #x=5#.

First of all, we can divide both members by #2#:

#(2x^2-12x+10)/2 = 0/2#

which simplifies into

#x^2-6x+5=0#

Arrived at this point, we can complete the square:

#x^2-6x+9-9+5=0#

And we notice that #x^2-6x+9=(x-3)^2#.

So, the whole expression becomes

#(x-3)^2 -4 = 0#

#\iff#

#(x-3)^2 = 4#

#\iff#

#x-3 = \pm 2#

#iff#

#x=3\pm 2 \implies x=1# or #x=5#.