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

Question

2870 views around the world

You can reuse this answer
Creative Commons License

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:

$t = -1/11$

How do you solve 3t(t+5)-t^2=2t^2+4t-13t(t+5)-t^2=2t^2+4t-1?