# How do you solve the quadratic 3y^2-y-4=0 using any method?

Oct 7, 2016

$y = - 1 \mathmr{and} y = \frac{4}{3}$

#### Explanation:

Find factors of 3 and 4 which subtract to give 1.

Notice that the numbers themselves are the required factors because $4 - 3 = 1$

$\left(3 y - 4\right) \left(y + 1\right) = 0$

Either of the 2 brackets could be equal to 1.

If $3 y - 4 = 0 \rightarrow y = \frac{4}{3}$

If $y + 1 = 0 \rightarrow y = - 1$