How to solve a complete quadratic equation in three ways

can you solve me this complete quadratic equation using by factoring and by completing the square in this problem?

3x^2-2x-5=0

Here's how to solve it by factoring:
3x² - 2x - 5 = 0
(3x - 5)(x + 1) = 0
3x - 5 = 0, x + 1 = 0
3x = 5, x = -1
x = 5/3, x = -1

And here's how to do it by completing the square:
3x² - 2x - 5 = 0
x² - (2/3)x - 5/3 = 0
x² - (2/3)x = 5/3
x² - (2/3)x + 1/9 = 5/3 + 1/9
(x - 1/3)² = 16/9
x - 1/3 = 4/3, x - 1/3 = -4/3
x = 5/3, x = -1

Hope that helps :)

Factoring
3x^2-2x-5=0
(x+1)(3x-5)=0
x+1=0; 3x-5=0
x=-1; x=5/3

Completing the Square:
3x^2-2x-5=0
3x^2-2x=5
3(x^2-2x/3)=5
3(x^2-2x/3+1/9)=44/9
3((3x-1)^2/9)=44/9
3(3x-1)^2=44
3(3x-1)=±2√11
3x-1=±(2√11)/3
3x=-3; 3x=5
x=-1; x=5/3

Quadratic Formula:
3x^2-2x-5=0
x=(2±√-2^2-4(3)(-5))/2(3)
x=(2±√4+60)/6
x=(2±8)/6
x=-1; x=5/3

X-5=0 x+1=0 therefore x=5 and x=-1 :-)