Help Please with Math (last question.) PLEASE?

Question

1. x^2 + 8x + 10=0 ; completing the square
A.{ -4 +- 2√12}
B.{-4 +- 2√6}
C.{-4 +- √6}
D.{4 +- √12}

2. 4 (z-3)^2 = 36
A.{-3,6}
B.{0,6}
C.{-9,9}
D.{6}

3. 6x^2 + 12x =0
A. {-6,-2}
B. {0,1/2}
C. {0,2}
D. {0,-2}

Answer 1

1.x^2+8x= -10
=>(x)^2+2*x*4+(4)^2=-10+16[adding 4^2 or 16 to both sides]
=>(x+4)^2=6
=>x+4=+-sqrt6 [square rooting both sides]
=>x= -4+-sqrt6
Hence C is the correct answer
2.4(z-3)^2=36
=>(z-3)^2=9[sividing both sides by 4]
=>z-3=+-3[square rooting both sides]
z=3+-3=6 or 0
B is the right answer
3. 6x^2+12x=0
=>6x(x+2)=0
Therefore either 6x=0 i.e. x=0
or x+2=0 i.e. x= -2
x=0 or -2
D is the correct answer

Answer 2

1)ANSWER IS B
2)
(z-3)^2=36/4
(z-3)^2=9 now use the formula (a-b)^2=a^2-2ab+b^2
z^2 - 6z + 9=9
z^2-6z+9-9=0
z^2-6z=0
z(z-6)=0
first solution is z=0
second solution is z=6 so the correct answer is B

3)
6x(x+2)=0
6x=0
x=0
and
x+2=0
x= -2 so the answer is D

Answer 3

x = (-8 +/- sqrt(64 - 40))/2 =
(-8 +/- sqrt(24))/2 =
(-8 +/- 2 * sqrt(6))/2 =
-4 +/- sqrt(6)

4(z-3)^2 = 36
(z-3)^2 = 9
z - 3 = +/- 3
z = 3 +/- 3
z = 6 or 0

6x(x + 2) = 0
x = 0 or x = -2