How do you solve #5t^2=25t#?
1 Answer
Jan 12, 2017
Explanation:
Collect terms on left side of equation thus equating to zero.
subtract 25t from both sides.
#5t^2-25t=cancel(25t)cancel(-25t)#
#rArr5t^2-25t=0# Take out a common factor of 5t
#5t(t-5)=0# We now have a product of factors equal to zero.
#"Thus " 5t=0rArrt=0#
#"or " t-5=0rArrt=5#
