use the quadratic formula to solve the equations.
2-2x=3x^2
2 - 2x = 3x^2
3x^2 + 2x - 2 = 0
x = [- b ± √(b^2 - 4ac)]/2a
a = 3, b = 2, c = - 2
there you go...make a contribution...you can do it.
@ß
[x = [- 2 ± 2√7]/6 = [- 1 ± √7]/3]
Rearrange to 3x^2+2x-2=0
This then fits the model ax^2+bx+c=0
So a=3, b=2, c=-2.
Substitute into the quadratic formula which is
x= [-b +/- sqr rt (b^2 - 4ac)]/2a
so you get x = [-2 +/- sqr rt (2^2 -4x3x-2)]/(2x3)
x = (-2 +/- 5.29)/6
so x = 0.549 or x = -0.8819
Comments
use the quadratic formula to solve the equations.
2 - 2x = 3x^2
3x^2 + 2x - 2 = 0
x = [- b ± √(b^2 - 4ac)]/2a
a = 3, b = 2, c = - 2
there you go...make a contribution...you can do it.
@ß
[x = [- 2 ± 2√7]/6 = [- 1 ± √7]/3]
Rearrange to 3x^2+2x-2=0
This then fits the model ax^2+bx+c=0
So a=3, b=2, c=-2.
Substitute into the quadratic formula which is
x= [-b +/- sqr rt (b^2 - 4ac)]/2a
so you get x = [-2 +/- sqr rt (2^2 -4x3x-2)]/(2x3)
x = (-2 +/- 5.29)/6
so x = 0.549 or x = -0.8819