How do you solve #x^2+4x=21#?

Question

1124 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-08T21:12:40

#x=-7# or #x=3#

let #x^2 + 4x# = 21
Moving the 21 to the other side we get #x^2 +4x -21 = 0#
Now, we want two numbers that multiply to give -21 and and to give 4

These two numbers are 7 and -3.
7 x -3 = 21
7+-3 = 4

Factorizing this we get:

#(x+7)(x-3)#

Now we make both brackets equal to zero
#x+7# = 0 or #x-3#=0
#x# = -7 #x#=3

Answer 2 · 2017-04-08T21:13:01

See below.

We move everything onto one side, and factor.

#x^2+4x-21=0#

#x^2-3x+7x-21=0#

#x(x-3)+7(x-3)=0#

#(x+7)(x-3)=0#

#x=-7,3#