How do you solve $y^{2} + y = 20$ $y^2+y=20$ ?

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Answer 1 · 2017-12-21T01:38:14

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

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

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

$+ 5$ $+5$ and $- 4$ $-4$ .

Therefore,

$(y + 5) (y - 4) = 20$ $(y+5)(y-4) = 20$

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

Answer 2 · 2017-12-21T01:38:24

$y = - 5, 4$ $y = -5, 4$

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

Subtracting $20$ $20$ From Both Sides,

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

Factoring,

$(y + 5) (y - 4) = 0$ $(y+5)(y-4) = 0$ .

Since One Of The Answers Have To Be 0,

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

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

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

How do you solve y2+y=20y^2+y=20?