Algebra 2: Solve ax^2 + bx + c = 0 by Factoring .... Homework HELP !?

0 = 12r^2 - 38r + 16 /:2

=> 0 = 6r^2 - 19r + 8

=> 0 = (2r - 1)*(3r - 8)

=> r1 = 1/2; r2 = 8/3

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First get everything on one side of the equation:

12r^2 - 38r + 16 = =0

Then, factor out a 2:

2 (6r^2 - 19r + 8) = 0

then, just factor by guess and check:

2 (2r - 1) ( 3r - 8) = 0

2 cannot equal 0, so 2r - 1 = 0 AND 3r - 8 = 0

2r - 1 = 0

2r = 1

r = 1/2

3r - 8 = 0

3r = 8

r = 8/3

so r = 1/2 and 8/3

16 = 38r - 12r^2

12r^2 - 38r + 16 = 0

12r^2 - 32r - 6r + 16 = 0

4r(3r - 8) - 2(3r - 8) = 0

(4r - 2)(3r - 8) = 0

2(r - 1)(3r - 8) = 0

r - 1 = 0 or 3r - 8 = 0

r = 1 or 3r = 8

r = 1 or r = 8/3 =~ 2.666666666666...

Having somebody else do them wont help on your attempt. indexed here are the 1st 2 to get an theory of what it needs to look as though. attempt something on your individual. 3x^2 + 2x - 8 (3x-4)(x+2) 2a^ + 5a + 3 (2a+3)(a+a million)

a= -12

b= 38

c= -16

Well I'm not very good in maths..