Algebra

[erobin8]

My problem is 3(x-5)^2=15. I chose this as my next step (3x-15)^2=15. Then (3x-15)(3x-15)=15. This gives me 9x^2-90x+225=15. From here, I don't know whether to set it to zero or subtract 225 from 15. Please help.

 

[Deleted member 4993]

My problem is 3(x-5)^2=15.

You do not say - what you need to do. I presume you need to solve for 'x'.

I chose this as my next step (3x-15)^2=15. That is incorrect.

Then (3x-15)(3x-15)=15. This gives me 9x^2-90x+225=15. From here, I don't know whether to set it to zero or subtract 225 from 15. Please help.

3(x-5)^2=15

(x-5)^2 = 5

x^2 - 10x + 25 = 5

x^2 - 10x + 20 = 0

Now solve the quadratic equation....