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

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Answer 1 · 2016-05-06T16:32:46

Expand and simplify to find:

$t = - \frac{1}{11}$ $t=-1/11$

Expanding the left hand side we find:

$3 t (t + 5) - t^{2} = 3 t^{2} + 15 t - t^{2} = 2 t^{2} + 15 t$ $3t(t+5)-t^2 = 3t^2+15t-t^2 = 2t^2+15t$

So the equation becomes:

$2 t^{2} + 15 t = 2 t^{2} + 4 t - 1$ $2t^2+15t = 2t^2+4t-1$

Subtract $2 t^{2}$ $2t^2$ from both sides to get:

$15 t = 4 t - 1$ $15t = 4t-1$

Subtract $4 t$ $4t$ from both sides to get:

$11 t = - 1$ $11t = -1$

Divide both sides by $11$ $11$ to get:

$t = - \frac{1}{11}$ $t = -1/11$

How do you solve 3t(t+5)−t2=2t2+4t−13t(t+5)-t^2=2t^2+4t-1?