3x^2 - 7x + 2 = 0?

Method 1
(3x - 1)(x - 2) = 0
x = 1 / 3 , x = 2

Method 2
x = [ - b ± √ (b ² - 4 a c ) ] / 2 a
x = [ 7 ± √ (49 - 24 ) ] / 6
x = [ 7 ± 5 ] / 6
x = 2 , x = 1 / 3

Try to factor it. If you can't factor it, use quadratic formula.

3x^2 - 7x + 2 = 0
(3x - 1)(x - 2) = 0

(3x - 1) = 0 or (x - 2) = 0

x = 1/3 or x = 2

Luckily you can solve this very easily by factorisation:

3x^2 - 7x + 2 = 0
3x^2 - 6x - x + 2 = 0
3x(x - 2) - 1(x - 1) = 0
(3x - 1)(x - 2) = 0

Either 3x - 1 = 0 (or) x - 2 = 0
x = 1/3 (or) x = 2

2 and 1/3 are the roots of the given equation

3x^2 - 7x +2 = 0

by factoring

(3x-1)(x-2)=0
1st factor
(3x-1)=0
3x=1
x=1/3.......^^,
2nd factor
(x-2)=0
x=2.........^^,

by quadratic formula

3x^2 - 7x +2 = 0........a^2 x+ bx + c = 0

the formulas are....
{-b + (b^2 - 4ac)^(1/2)} / 2a & {-b - (b^2 - 4ac)^(1/2)} / 2a

x = {-b + (b^2 - 4ac)^(1/2)} / 2a
= { -(-7) + [(-7)^2 - 4(3)(2)]^(1/2) } / 2(3)
= { 7 + [49 - 24]^(1/2) } / 6
= { 7 + 25^(1/2) } / 6
= { 7 + 5 } / 6
= 12/6
=2.............x=2

x = {-b - (b^2 - 4ac)^(1/2)} / 2a
= { -(-7) - [(-7)^2 - 4(3)(2)]^(1/2) } / 2(3)
= { 7 - [49 - 24]^(1/2) } / 6
= { 7 - 25^(1/2) } / 6
= { 7 - 5 } / 6
= 2/6
=1/3.............x=1/3

3x^2 - 7x + 2 = 0
(3x - 1)(x - 2) = 0

(3x - 1) = 0 ---> x = 1/3
(x - 2) = 0 ---> x = 2

Therefore, x = 1/3 and x = 2 are your solutions.

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

this can be factorised into (3x - 1)(x - 2) = 0
(multipliying this out gives 3x² - x - 2(3x) + 2 = 3x² - 7x +2)

so either bracket must equal 0:
3x - 1 = 0 => 3x = 1 => x = 1/3
or x - 2 = 0 => x = 2

these are your two solutions

(3x-1)(x-2)=0
x=1/3 or 2