How do you solve #x^2+4x=21#?
Moving the 21 to the other side we get
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:
Now we make both brackets equal to zero
We move everything onto one side, and factor.