計極都計唔到

3 ( x^2-1) = 2x/3

Multiply both sides by 3,

9 (x^2 - 1) = 2x
9x^2 - 9 = 2x
9x^2 - 2x - 9 = 0

By using the quadratic formula,
x = {-(-2) ± √[(-2)^2-4(9)(-9)] } / [(2)(9)]
= {2 ± √(4 + 324)} / 18
= [2 ± √(328)] / 18
= [2 ± 2√(82)] / 18
= [1 ± √(82)] / 9

If there is a mistake, please inform me.

3(x^2-1)=2x/3

3x^2-3=2x/3

3(3x^2-3)=2x

9x^2-2x-9=0

∴ {2±√[(-2)^2-(4)(9)(-9)}/2*9

=[2±√(4+324)]/18

=(2±√328)/18

=(2±2√82)/18

=(1±√82)/18

3 ( x^2-1) = 2x/3
=> 9 (x^2 - 1) = 2x by multiplying 3 on both sides
=> 9x^2 - 2x - 9 = 0 by adding 2x on both sides
Then, discriminant, d = [(-2)^2 - 4(9)(-9)]^(1/2) = [2^2 x (1 + 81)]^(1/2) = 2x(82)^(1/2)
By quadratic formula,
x = [-(-2) + d]/(2x9) or [-(-2) - d]/(2x9) = [1 + 82^(1/2)]/9 or [1 - 82^(1/2)]/9

3 ( x^2-1) = 2x/3

3x^2 - 3 = 2x/3

3 (3x^2 - 3) = 2x

9x^2 - 9 = 2x

9x^2 - 2x -9 = 0

x= [-b + √(b^2 -4ac)] / 2a

x=2 + √[4 - 4(9)(-9)]/2(9)

x=(2+ √328)/18 or x = (2- √328)/18

3 ( x^2-1) = 2x/3

=> 9x^2 - 9 = 2x

=> 9x^2 - 2x - 9 = 0

using quadratic formula,
x = (2 +/- sqrt(328)) / 18 = (2 +/- 2 * sqrt(82)) / 18 = (1 +/- sqrt(82)) / 9