Need help solving these equations.... (algebra 1)

I'm studying for finals, and in the practice test I have, I have coming across some difficulties with these (types) of problems. Your help and explanations would be much appreciated :)

I have to solve the equations and express these in the simplest form:

1. (x+4)^2=28

2. x^2=7-8x

You have to factor them:

(x+4)(x+4)=28 >because it was to the power od 2.

F.O.I.L.

x^2+4x+4x+16=0 -simplify = x^2+8x+16

factor your equation

(x+4)(x+4)

set these to 0

x+4=0 x+4=0

solve for x

x+4=0 x+4=0
-4 -4 -4 -4

x= -4 x= -4


Do the same thing for the next equation but subtract the x^2. And after you do that the x^2 will be negative and to get rid of that divide the whole equation by -1. Good Luck!

1) use the distributive property to get:
x^2 + 4^2 = 28
use the substitution property (4^2 = 16) to get:
x^2 + 16 = 28
subtract 16 from both sides to get:
x^2 = 28 - 16 = 12
get the square root of both sides and you end up with:
x = the square root of 12

2) add 8x to both sides (canceling it out on one to side) to get:
xx + 8x = 7

(the xx = x times x = x^2)

xx + 8x = x(8+x)
so
x(8+x) = 7
that's the simplest it'll get


(unless you'd rather use: 8+x = 7/x)

1. (x+4)^2=28
or, x+4 = +/-sqrt(28)
or, x+4 = +/-2sqrt(7)
or, x = [2sqrt(7)-4] and [x=-2sqrt(7)-4]


2. x^2=7-8x
or, x^2+8x-7 = 0
or, x^2+2.x.4+4^2-4^2-7 = 0
or, (x+4)^2-23 = 0
or, x+4 = +/-sqrt23
or, x = [-sqrt(23)-4] and [sqrt(23)-4]

(x+4)^2 = 28
x+4 = ±√28
x = -4 ± √28
x= -4 ± 2√7


x^2 = 7 - 8x
x^2 + 8x = 7
x^2 + 8x + 16 = 7 + 16
(x+4)^2 = 23
x+4 = ±√23
x = -4 ±√23

What is ^? To the quantity squared?