write each quadratic equation in standard form then identify the value of a b and c

We want the form  ax^2 + bx + c = 0

1. 3x - 2x^2  = 7   re-write as

-2x^2 + 3x - 7  = 0

a = -2, b = 3, c = -7

2.  5 - 2x^2  = 6x    re-write as

-2x^2 - 6x + 5 = 0

a = -2, b = -6, c = 5

3. (x + 3) ( x + 4)  = 0   expand as

x^2 + 3x + 4x + 12 = 0

x^2   +  7x + 12  = 0

a =1. b = 7, c  = 12

4. (2x + 7) ( x+ 4)  = 0  expand as

2x^2 + 7x + 8x + 28   =  0

2x^2  + 15x + 28  = 0

a = 2, b = 15, c = 28

5.  2x ( x - 3)   = 15

2x^2 - 6x  = 15

2x^2 - 6x - 15 = 0

a = 2, b = -6, -15

and there's more 

1.)(x+7)(x-7)=-3x

2.)(x-4)^2+8=0

3.)(x+2)^2=3(x+2)

4.)(2x-1)^2=(x+1)^2

5.)2x(x+4)=(x-3)(x-3)

thank you cphill god bless you

1. (x + 7) ( x - 7)  = -3x

x^2 - 49 = -3x

x^2 + 3x - 49 = 0

a = 1, b = 3, c = -49

2. (x - 4)^2 + 8 = 0

(x - 4) ( x- 4) + 8 =0

x^2 - 4x - 4x + 16 + 8 =0

x^2 - 8x + 24 = 0

a = 1, b = -8, c = 24

3.  ( x + 2 )^2  = 3 ( x + 2)

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

(x + 2) ( x + 2) - 3(x + 2)  = 0

x^2 + 2x + 2x + 4 - 3x - 6  = 0

x^2 + 4x + 4 - 3x - 6  = 0

x^2 + x - 2  = 0

a = 1, b = 1, c = -2

4. (2x + 7) ( x + 4)   = 0

2x^2 + 7x + 8x + 28  = 0

2x^2 + 15x + 28  = 0

a = 2 , b =  15, c = 28

5. 2x( x + 4)  =  (x -3) ( x - 3)

2x^2 + 8x  - (x - 3) ( x - 3)  = 0

2x^2 + 8x -  [ x^2 - 6x + 9 ]  = 0

2x^2 + 8x - x^2 + 6x - 9  = 0

x^2 + 14x - 9  = 0

a = 1, b = 14, c = -9

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

(2x - 1) ( 2x - 1)  =  3( x + 1) ( x + 1)

4x^2 - 2x - 2x + 1  =  3 ( x^2 + 1x + 1x + 1)

4x^2 - 4x + 1  =  3 ( x^2 + 2x + 1)

4x^2 - 4x + 1  =  3x^2 + 6x + 3   simplify

x^2 - 10x - 2  =  0

a = 1,  b = -10,  c = -2