How do you solve #5y^2=30y#?

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Answer 1 · 2018-07-14T16:12:20

#y=0# or #y=6#

Writing your equation in the form
#5y(y-6)=0#
so we get

#y=0# or #y=6#

Answer 2 · 2018-07-14T16:14:14

#y=0# and #y=6#

We have a second-degree term, so this is beginning to look like a quadratic. Let's subtract #30y# from both sides to find the zeroes:

#5y^2-30y=0#

We can factor a #5y# out, and we get

#5y(y-6)=0#

Setting both factors equal to zero, we get

#y=0# and #y=6#

Hope this helps!