SOLUTION: I am having trouble solving this problem. It is in the book Introductory Algebra. 12/sqrt5x+6 = sqrt2x+5 Here is what I have so far. (12/sqrt5x+2)^2= (sqrt2x+5)^2 144/5

Algebra ->  Radicals -> SOLUTION: I am having trouble solving this problem. It is in the book Introductory Algebra. 12/sqrt5x+6 = sqrt2x+5 Here is what I have so far. (12/sqrt5x+2)^2= (sqrt2x+5)^2 144/5      Log On


   

Question 275967: I am having trouble solving this problem. It is in the book Introductory Algebra.

12/sqrt5x+6 = sqrt2x+5
Here is what I have so far.

(12/sqrt5x+2)^2= (sqrt2x+5)^2
144/5x+6 =2x+5
(5x+6)(2x+5)=144
10x^2+37x+30=144
10x^2+37x-114=0
I can't find two solutions for 10x^2+37x-114 so I'm not sure what to do.
Answer by mananth(16946) About Me  (Show Source):
You can put this solution on YOUR website!
12/sqrt5x+6 = sqrt2x+5
144 / 5x+6 = 2x+5
(2x+5)(5x+6)= 144
10x^2+12x+25x+30=144
10x^2+37x-114=0
The factors are not integers
so you have to find the roots of the equation
x1,x2 = -b + / - {sqrt(b^2-4ac)} / 2a
a= 10 b=37 and c=-114
x1= -b+{sqrt(b^2-4ac)} / 2a
= -37+sqrt(1369+4560) / 20
x1=2
x2= -b-{sqrt(b^2-4ac)} / 2a
x2= -114/20
x2= -5.7
2 & -5.7 are the roots of the equation