Asked by Anonymous

Solve for x.

x(x+3)=2^2

I think I'm doing this all wrong. Here is the work I've done so far:

x^2+3x=4
x^2+3x-4=0

...now what?
i started trying to use the equation x=-b sqrt of (b^2-4ac)
but I'm just getting confused...
Answered by bobpursley
repeat
Answered by Damon
x^2+3x-4=0 OK, now:

(x+4)(x-1) = 0
x = 1 or x = -4

or
x = [-3 +/- sqrt (9+16) ] / 2
= -3/2 +/- 5/2
= -1.5 +/- 2.5
= -4 or +1 again