How do you solve # x^2+4x-5=0 #?

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Answer 1 · 2018-03-07T01:32:19

#x= -5 or +1#

First, factor this equation. Then, determine what two values add to equal to the middle term #(+4)# and multiply to the last term #(-5)#,

#+5 and -1#

Plug it in the equation:

#(x+5)(x-1)#

and set equal to zero

#(x+5)(x-1)=0#

#x+5=0 => x=-5 #

#x-1=0 => x=+1#