math help please?!?!?!?!?! 10 points for best answer!!!!!!?

Question

1. Solve the Quadtratic equation:
3x^2 - 8x + 1 = 0

is it:
[a] -4 +or- 2sqrt(13) /3
[b] -4 +or- sqrt(13) /3
[c] 4 +or- 2sqrt(13) /3
[d] 4 +or- sqrt(13) /3


2. 2x^2 - 1 = 6x

is it:
[a] -3 +or- sqrt(11) /2
[b] 3 +or- sqrt(11) /2
[c] -3 +or- sqrt(37) /2
[d] 3 +or- sqrt(7) /2




please show me how to do it
thanks :]

Answer 1

3x^2 - 8x + 1 = 0
x= [-(-8) +/- sqrt(8^2-4(3)(11)]/(2*3)
x = [8 +/-sqrt(52)]/6
x =[8 +/- 2sqrt(13)]/6
x = [4 +/- sqrt(13)]/3

2. 2x^2 - 1 = 6x
2x^2-6x-1=0
x = [-(-6) +/- sqrt(6^2 -4(2)(-1)]/(2*2)
x = [6 +/- sqrt(44)]/4
x = [6 +/- 2sqrt(11)]/4
x = [3 +/- sqrt(11)]/2

Answer 2

The solved answer is:
1.[d] 2.[b]

The how to do it is this:

First adjust your eq to solve so on the right side has nothing but 0
(Ex: eq1:::::3x^2 - 8x + 1 = 0, eq2::::: 2x^2-6x-1=0)

Copy the numbers "a", "b" and "c" on your eq to solve and paste them on this formula:

X=[b-(b^2±4ac)^0.5]/2a
(± symbol means: you first add then substrac. So you'll have two results the one you add and the one you substrac)

The numbers are those:
a: is the one which multiplies x^2 (Ex: 3 on eq1, 2 on eq2)
b: is the one which muliplies x (Ex: -8 on eq1, -6 on eq2)
c: is the constant (Ex: 1 on eq1, -1 on eq2)

Answer 3

Use the quadratic equation. x=[ -b +- sqrt(b^2 - 4ac)]/2a where ax^2 + bx + c=0.

Answer 4

im not that smart