How do you solve #7x^2+10x=2x^2+155# using any method?

Question

1284 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-12T20:08:09

#x=4sqrt2-1# or #x=4sqrt2+1#

#7x^2+10x=2x^2+155#

#hArr7x^2+10x-2x^2-155=0# or

#5x^2+10x-155=0# and dividing by #5#

#x^2+2x-31=0#

and using quadratic formula #(-b+-sqrt(b^2-4ac))/2a# , as we have here #a=1#, #b=2# and #c=-31#

#x=(-2+-sqrt((-2)^2-4×1×(-31)))/(2×1)# or

#x=(-2+-sqrt(4+124))/(2×1)# or

#x=(-2+-sqrt128)/2# or

#x=(-2+-8sqrt2)/2=-1+-4sqrt2# or

#x=4sqrt2-1# or #x=4sqrt2+1#