(3x+2)*(x-1)=100?

(3x + 2)(x - 1) = 100
3x² - x - 2 = 100
3x² - x - 102 = 0
x = [ 1 ± √ (1 + 1224 ) ] / 6
x = [ 1 ± √ (1225 ) ] / 6
x = [ 1 ± 35 ] / 6
x = 6 , x = - 34/6
x = 6 , x = - 17/3

[1]
(3x+2)(x-1)=100
3x^2-x-2=100
3x^2-x-2-100=0
3x^2-x-102=0
Comparing the equation with standard quadratic equation ax^2+bx+c=0,we get
a=3,b= -1 and c= -102
x={-b+-sqrt(b^2-4ac)}/2a
={1+-sqrt(1+1224)}/6
=(1+sqrt1225)/6
=(1+-35)/6
=36/6 or -34/6
=6 or -17/3
Therefore x=6 or -17/3 ans

(3x + 2)(x - 1) = 100
3x^2 + 2x - 3x - 2 = 100
3x^2 - x - 2 - 100 = 0
3x^2 - x - 102 = 0
(3x + 17)(x - 6) = 0

3x + 17 = 0
3x = -17
x = -17/3
x = -5 2/3

x - 6 = 0
x = 6

∴ x = -5 2/3 , 6

foil (3x+2)*(x-1)=3x^2-x-2=100

3x^2-x-2-100 = 0
3x^2-x-102=0 factor
(x-6)(3x+17)=0
x-6 = 0 or (3x+17)=0
x = 6 or x = -17/3

(x-1) = 100
x = 101

(3x+2) = 100
x = (100 - 2)/3

(3x+2)*(x-1)=100
3x^2 - x - 2 = 100
3x^2 - x - 102 = 0
(3x + 17)(x - 6) = 0

3x + 17 = 0
x = -17/3

x - 6 = 0
x = 6

x = {-17/3, 6}