i keep getting the wrong answers everytime i did these two problems, someone please solve them for me.?

Question

i don't get quadratic formulas:

x(squared)+3x-3=0

3x(squared)-7x=12

Answer 1

x^2 + 3x - 3 = 0
x^2 + 3x = 3
add 9/4 both sides
x^2 + 2(3/2)x + 9/4 = 3+ 9/4
(x + 3/2)^2 = 21/4
x + 3/2 = + or -sqrt(21/4)
x + 3/2 = + or - sqrt(21)/2
x = -3/2 + sqrt(21)/2 or -3/2 - sqrt(21)/2
x = (1/2)[-3 + (21)^(1/2)] or -1/2[(3 + (21)^(1/2)]

2)
3x^2 - 7x = 12
divide by 3
x^2 - 7/3x = 4
add 49/36 both sides
x^2 - 2(7/6)x + 49/36 = 49/36 + 4
[x - (7/6)]^2 = 189/36
= 9*21/36
= [(3/6)sqrt(21)]^2


x - 7/6 = + or - (3/6) sqrt(21)
x = 7/6 + (3/6)sqrt(21) or 7/6 - (3/6)sqrt(21)
= (1/6)(7 + 3 sqrt(21) or (1/6)[7 - 3 sqrt(21)]

Answer 2

x =[-b ±√b²-4ac] / 2a

a = 1
b = 3
c = -3

and solve

2nd one

a = 3
b = -7
c = -12

and solve