Posted by **Ana** on Friday, September 3, 2010 at 8:55pm.

how do i solve for y in the equation -5y(6y - 6) - y = -4(y - 5)

- 10th grade math -
**Reiny**, Friday, September 3, 2010 at 9:25pm
-5y(6y - 6) - y = -4(y - 5)

-30y^2 + 30y - y = -4y + 20

30y^2 - 33y + 20 = 0

y = (22 ± √-1311)/60

or

(22 ± i √1311)/60 , two complex roots.

- 10th grade math -
**danny**, Monday, May 23, 2011 at 8:27pm
how kan i solve this equation

7a-3a+2a-a=16

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