SOLUTION: 2x^2-5=-55 2x^2-50=0 2(x^2-25)=0 2(x-5)=0 2(x+5)=0 x=5 x=5 I checked my work for both x's and it does not come out right for both I come up with 45=55. I am not wure what

Algebra ->  Radicals -> SOLUTION: 2x^2-5=-55 2x^2-50=0 2(x^2-25)=0 2(x-5)=0 2(x+5)=0 x=5 x=5 I checked my work for both x's and it does not come out right for both I come up with 45=55. I am not wure what      Log On


   



Question 400295: 2x^2-5=-55
2x^2-50=0
2(x^2-25)=0
2(x-5)=0 2(x+5)=0
x=5 x=5
I checked my work for both x's and it does not come out right for both I come up with 45=55. I am not wure what I am doing wrong here. What is the answer to x? Please show all work and how to check the answer.

Answer by scott8148(6628) About Me  (Show Source):
You can put this solution on YOUR website!
adding 5 ___ 2x^2 = -50

dividing by 2 ___ x^2 = -25

there are no REAL numbers that you can square and get a negative value

"i" is the notation used for the square root of -1

x^2 = 25(-1) = 25 i^2

taking square root ___ x = ±5i