# How do you solve y^2+y=20?

Dec 21, 2017

${y}^{2} + y = 20$

${y}^{2} + y - 20 = 0$

What $2$ numbers add to $+ 1$ and multiply to $- 20$?

$+ 5$ and $- 4$.

Therefore,

$\left(y + 5\right) \left(y - 4\right) = 20$

Therefore, $y$ can be color(red)(-5 or, +4

Dec 21, 2017

$y = - 5 , 4$

#### Explanation:

${y}^{2} + y = 20$.

Subtracting $20$ From Both Sides,

${y}^{2} + y - 20 = 0$.

Factoring,

$\left(y + 5\right) \left(y - 4\right) = 0$.

Since One Of The Answers Have To Be 0,

$y + 5 = 0$, $y = - 5$

$y - 4 = 0$, $y = 4$

The Answers Are: $- 5$ and $4$.