solve the quadratic equation 12x^2= 3x+9?

12x^2= 3x+9
12x^2 -3x-9 =0
divide by 3
4x^2 -x-3 =0
factorize
(4x+3)(x-1) =0
4x+3 =0 => x= -3/4;
x-1=0 => x= 1;

Roots are x= -3/4 or x= 1;

Hope this helps
Bye
gilvi

12x² - 3x = 9
x² - 1/4x = 3/4
x² - 1/8x = 3/4 + (1/8)²
x² - 1/8x = 48/64 + 1/64
(x - 1/8)² = 49/64
x - 1/8 = 7/8

1st factor:
= x - 1/8 + 7/8
= x + 3/4
= 4x + 3

2nd factor:
= (12x² - 3x - 9)/(4x + 3)
= 3x - 3 or 3(x - 1)

Values of x:
4x + 3 = 0, 4x = - 3, x = - 3/4
x - 1 = 0, x = 1

Answer: x = - 3/4, 1

Proof (where x = - 3/4):
12(- 3/4)² = 3(- 3/4) + 9
12(9/16) = - 9/4 + 36/4
27/4 = 27/4

Proof (where x = 1):
12(1²) = 3(1) + 9
12(1) = 3 + 9
12 = 12

4x² - x - 3 = 0
(4x + 3)(x - 1) = 0
x = - 3/4 , x = 1

12x^2= 3x+9
12x^2- 3x-9 = 0
factor out a 3

3(4x^2 -1x -3) =0
3(4x +3)(x -1)=0

by the zero property
each bracket equals zero
(x -1)=0 or (4x +3) =0
now solve for x in both cases
x = 0 or x = -3/4

12x^2 = 3x + 9
12x^2 - 3x - 9 = 0
simplified:-
4x^2 - x - 3 = 0
(4x + 1)(x - 3) = 0
x = -1/4 OR x = 3

12x^2 = 3x + 9
12x^2 - 3x - 9 = 0
(12x^2 - 3x - 9)/3 = 0/3
4x^2 - x - 3 = 0
4x^2 + 3x - 4x - 3 = 0
(4x^2 + 3x) - (4x + 3) = 0
x(4x + 3) - 1(4x + 3) = 0
(4x + 3)(x - 1) = 0

4x + 3 = 0
4x = -3
x = -3/4 (-0.75)

x - 1 = 0
x = 1

∴ x = -3/4 (-0.75) , 1

12x^2= 3x+9 -----12x^2-3x-9=0
D = 9+4*12*9 = 441

x = (3+21)/24 = 24/24 = 1
x = (3-21)/24 = -18/24 = -3/4 =-0.75

12x^2 - 3x - 9 = 0
(4x + 3)(3x - 3) = 0
x = -3/4 or x = 1