How do you solve 6t ^ { 2} = - 4t?

Apr 5, 2018

$t = 0 \mathmr{and} t = - \frac{2}{3}$

Explanation:

$6 {t}^{2} + 4 t = 0$

$2 t \left(3 t + 2\right)$=0

$2 t = 0 \mathmr{and} 3 t = - 2$

$t = 0 \mathmr{and} - \frac{2}{3}$

Apr 5, 2018

$t = 0$ and $t = - \frac{2}{3}$

Explanation:

Add $- 4 t$ to both sides:

$6 {t}^{2} + 4 t = \cancel{- 4 t + 4 t}$

$6 {t}^{2} + 4 t = 0$

We can then factor out $2 t$:

$2 t \left(3 t + 2\right) = 0$

We can then use the zero factor principle to know that this equation holds either when $2 t = 0$ or when $3 t + 2 = 0$. This gives us two different solutions:

$2 t = 0 \to t = 0$

$3 t + 2 = 0 \to - \frac{2}{3}$

These are the solutions to the equation.

Apr 5, 2018

The answer is : -2/3

Explanation:

Hello there!!

Let me explain how to do this.....
It is very simple.

Take,

$6 {t}^{2} = - 4 t$

Now " t " is cancelled out on both sides,

$6 t = - 4$

then,
$t = - \frac{4}{6}$

$t = - \frac{2}{3}$ or $t = 0$