algebra help?

Question

how to solve using the quadratic formula
6x^2 - 4x - 1 = 0

Answer 1

using the formula (I assume you know what it is) 6 is a, -4 is b and -1 is c. replace the letters with numbers... also i think you have to give two answers with the quadratic formula. You get a positive and a negative because with SQRT you just do.

Answer 2

6x^2 - 4x - 1 = 0

a = 6,b = - 4, c = -1
x = - b + - √ (b^2 - 4ac) / 2a
= - * - 4 + - √ [(-4)^2 - 4 * 6 * -1 ) / 2*6
= 4 + - √ ( 16 + 24 ) / 12
= 4 + - √ 40 / 12
= 4 + - √ (4 * 10 ) / 12
= 4 + - 2 √ 10 / 12

so x =
4 + 2 √ 10 / 12 or 4 - 2 √ 10 / 12
2( 2 + √10) / 12 ; 2( 2 - √10) / 12
(2 + √10) / 6 ; (2 - √10) / 6

so x = {(2 + √10) / 6 ; (2 - √10) / 6}

Answer 3

6x^2 - 4x - 1 = 0
x=(4±√(4²+4*6*1))/12
x=(4±√(16+24))/12
x=(4±√40)/12
x=(4±1√10)/12
x=1/3 ±1/6 √10

Answer 4

x = [4 ± √(16 + 24)] / 12
x = [4 ± √ 40 ] / 12
x = [4 ± 2√10] / 12
x = [2 ± √10] / 6
x = 0.86 , x = - 0.194

Answer 5

a=6, b=-4, c=-1

[-(-4) +/-sqrt((-4)^2-4(6)(-1))]/2(6)
[4+/-sqrt(40)]/12
this reduces to
1/3 +/-sqrt(10)/6