Please Help... Alg 2?!?

Question

How do I Solve the following equation for x by using the quadratic formula:

2x^2 - 6x - 2 = 0

(If there is more than one solution, separate them with commas.)

Thanks

Answer 1

x ² - 3x - 1 = 0
x = [ 3 ± √(9 + 4) ] / 2
x = [3 ± √(13)] / 2
x = [ 3 + √13 ] / 2 , x = [ 3 - √(13) ] / 2

Answer 2

Quadratic Formula

x = - b ± √b² - 4ac / 2a

2x² - 6x - 2 = 0

let

a = 2

b = - 6

c = - 2

x = - (- 6) ± √( - 6) ²- 4(2)(- 2) / 2(2)

x = 6 ± √36 - (8)(- 2) / 4

x = 6 ± √36 - (- 16) / 4

x = 6 ± √36 + 16 / 4

x = 6 ± √52 / 4

x = 6 ± 7.211102551 / 4

- - - - - - - - -

Solving for +

x = 6 + 7.211102551 / 4

x = 13.21110255 / 4

x = 3.302775638

- - - - - - -

Solving for -

x = 6 - 7.211102551 / 4

x = - 1.211102551 / 4

x = - 0.302775638

- - - - - - - s-

Answer 3

the quadratic formula is

x = [-b (+/-) sqrt (b^2 - 4ac)] / (2a)

a = 2
b = -6
c = -2

x = {6 (+/-) sqrt [(-6)^2 - 4(2)(-2)] } / (2)(2)

x = -0.30277563773199464655961063373525, 3.3027756377319946465596106337352

Answer 4

In that equation:

a = 2
b = -6
c = -2

The quadratic formula is:

x = (-b +/- sqrt(b^2 - 4ac))/2a
x = (6 +/- sqrt(36 +16))/4
x = (6 +/- sqrt(52))/4
x = (6 +/- 7.21)/4
x = (6 + 7.21)/4, (6 - 7.21)/4
x = 3.3, -0.3

Answer 5

use your quadratic equation x= {-b +/- sqrt(b^2 -4ac)}/2a to find x values, plug back in to find your corresponding y values.