How do you solve #3t(t+5)-t^2=2t^2+4t-1#?

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Answer 1 · 2016-05-06T16:32:46

Expand and simplify to find:

#t=-1/11#

Expanding the left hand side we find:

#3t(t+5)-t^2 = 3t^2+15t-t^2 = 2t^2+15t#

So the equation becomes:

#2t^2+15t = 2t^2+4t-1#

Subtract #2t^2# from both sides to get:

#15t = 4t-1#

Subtract #4t# from both sides to get:

#11t = -1#

Divide both sides by #11# to get:

#t = -1/11#