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

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Answer 1 · 2017-12-21T01:38:14

#y^2 +y = 20#

#y^2 + y - 20 =0#

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

#+5# and #-4#.

Therefore,

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

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

Answer 2 · 2017-12-21T01:38:24

#y = -5, 4#

#y^2+y=20#.

Subtracting #20# From Both Sides,

#y^2+y-20 = 0#.

Factoring,

#(y+5)(y-4) = 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#.