I am lost.. Can someone please help me??

Question

1. Solve by completing the square:
x^2 - 6x -3 = 0

2. Solve using the quadratic formula:
3x^2 - 2x + 1 = 0

Answer 1

x^2 - 6x -3 = 0
First move the constant (-3) to the other side. Add 3 to each side.
x^2 - 6x = 3
Now take -6x and divide it by two, then square it. Add that to both sides. [ -6/2=3^2=9 ]
x^2 - 6x + 9 = 3+9
x^2 - 6x + 9 = 12
(x-3)^2=12
Take the square root of both sides
√(x-3)^2=√12
x-3=√12
x=3±√12 or 3±2√3

2]]
3x^2 - 2x + 1 = 0
quadratic formula is (-b±√b^2-4ac)/2a
Here, A=3, B= -2, and C=1
SO just plug it in.

[-(-2)±√(-2)^2-4(3)(1)]/2(3)
(2±√4-12)/6
(2±√-8)/6
(2±2i√2)/6
(1±i√2)/3

have you learned imaginary numbers?!

Answer 2

Completing the square

x² - 6x - 3 = 0

Transpose 3

x² - 6x - 3 + 3 = 0 + 3

x² - 6x = 3

x² - 6x + ____= 3+____

Divide 6 in half equals 3 and square 3 equals 9. Insert 9 into the blank section.

x² - 6x + 9 = 3 + 9

x² - 6x + 9 = 12

(x - 3)(x - 3) = 12

(x - 3)² = 12

(√x - 3)² = ± √12

x - 3 = ± √12

Transpose 3

x - 3 + = 3 ± √12

x = 3 ± √12

x = 3 ± 3.464101615

- - - - - - - -

Solving for +

x = 3 + 3.464101615

X = 6.464101615

- - - - - - - - -

Solving for -

x = 3 - 3.363101615

x = 0.464101615

- - - - - - -s-

Answer 3

1) x=[3±√(9+3)] = 3+2√3 and 3-2√3

2) x=[1±√(1-3)]/3 = [1+i·√2]/3 and [1-i·√2]/3

Saludos.

Answer 4

1. x²-6x=3
x²-6x+9=3+9
(x-3)²=12
x-3=±2√3
x=3±2√3

2. x= -b±√(b²-4ac)
-------------------------
2a

x= 2±√(4-12)
-------------------
6

x= (2±2√2i)6 = (1±√2i)/3