# How do you solve 5t^2=25t?

Jan 12, 2017

$t = 0 , t = 5$

#### Explanation:

Collect terms on left side of equation thus equating to zero.

subtract 25t from both sides.

$5 {t}^{2} - 25 t = \cancel{25 t} \cancel{- 25 t}$

$\Rightarrow 5 {t}^{2} - 25 t = 0$

Take out a common factor of 5t

$5 t \left(t - 5\right) = 0$

We now have a product of factors equal to zero.

$\text{Thus } 5 t = 0 \Rightarrow t = 0$

$\text{or } t - 5 = 0 \Rightarrow t = 5$