## anonymous 5 years ago Solve for k: ((16-2k)/3)^2+(2((16-2k)/3))+k=40

1. anonymous

(256-64k+4k^2)/9 + (32-4k)/3 + k =40

2. anonymous

then put them all over 9

3. anonymous

(256-64k+4k^2 + 96-12k+9k-360)/9 = 0

4. anonymous

Wait how'd you get the first step?

5. anonymous

Never mind 16 squared?

6. anonymous

I expanded the first part squared and multiplied the second part by 2

7. anonymous

Do you know what to do from there?

8. anonymous

i think so

9. anonymous

just say so if you need more help

10. anonymous

okay i got to -64k+4k^2-12k+9k=-352, how do i get all the k's alone?

11. anonymous

you can group all the ks together and get 4k^2 + 67k +352 = 0

12. anonymous

then use the quadratic equation with a=4, b=67, c=352

13. anonymous

-67, sorry

14. anonymous

15. anonymous

$k=(-b \pm \sqrt{b ^{2}-4ac})/2a$

16. anonymous

gotcha!