algebra 2 help! please explain?

Question

-11x^2 + 12x - 6 = 0
A x = {-6 +- i[sqrt of 60)}/11
B x = {-6 +- i[sqrt(30)]}/-11
C x = {6 +- i[sqrt(12)]}/6
D x = {-6 +- i[sqrt(22)]}/-11
E x = {-11 +- i} /6
F x = {11 +- i} /-6

Answer 1

Quadtratic Equation: x = [-b +/- V`(b^2 - 4ac)] / 2a

First: you have three terms > a = -11; b = 12; c = -6
Sec: replace the terms with the corresponding variables.

x = [-12 +/- V`(12^2 - 4(-11)(-6))] / 2(-11)

x = [-12 +/- V`(12*12 - 4(-11)(-6))] / -22

x = [-12 +/- V`(144 - 4(66))] / -22

x = [-12 +/- V`(144 - 264)] / -22

x = [-12 +/- V`(-120)] / -22

Sec: rule - you can't find the square root of a negative number. The negative sign becomes an imaginary number known as "i" > express 120 in lowest terms.

x = [-12 +/- i V`(2*2*2*3*5)] / -22

x = [-12 +/- 2i V`(2*3*5)] / -22

x = [-12 +/- 2i V`(30)] / -22

x = (-12/-22) +/- [2i V`(30)] / -22

x = (-6/-11) +/- [i V`(30)] / -11

B. x = [(-6 +/- i V`(30)] / -11

P.S. the equation can be simplified...

x = [(6 +/- i V`(30)] / 11

Answer 2

Use the quadratic formula. If a quadratic equation is ax^2 + bx + c, you have a = -11, b = 12, and c = -6. The formula is x = (-b +/- sqrt(b^2 - 4ac)) / 2a. So x = (-12 +/- sqrt(12^2 - 4*(-11)*(-6))) / (2*(-11)) = (-12 +/- sqrt(144 - 264)) / (-22) = (-12 +/- sqrt(-120)) / (-22) = (-12 +/- 2*sqrt(30)) / (-22) = (-6 +/- sqrt(30)) / (-11), choice B. Personally, I would simplify this further to (6 +/- sqrt(30)) / 11

Answer 3

using the quadratic formula
x = (-12 +- (12^2 - 4(-11)(-6))^.5)/-22
x = -6/-11 +-(-120)^.5/-22
x = -6/-11 +- 2(-30)^.5/-22
x = -6/-11 +- i(30)^.5/-11
x = (-6 +- i(30)^.5)/-11