How do you solve #2y^2+6=y^2#?

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Answer 1 · 2015-05-23T15:19:55

#y=sqrt(6)i=isqrt6#

Problem: Solve #2y^2+6=y^2#.

Subtract #6# from both sides of the equation.

#2y^=y^2-6#

Subtract #y^2# from both sides.

#y^2=-6#

Take the square root of both sides.

#y=sqrt(-6)# =

#y=sqrt(6)i=isqrt6#