how do you solve the equation 12x^2+3x=0?

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

3x = 0
x = 0/3
x = 0

4x + 1 = 0
4x = -1
x = -1/4 (-0.25)

∴ x = -1/4 (-0.25), 0

Factor
3x(4x + 1) = 0
==> x = 0 or x = -1/4

3x ( 4x + 1 ) = 0

x = 0 , x = - 1/4

12x^2+3x=0

3x(4x+1)=0

x=0 or -1/4

stick the coefficients into te quadratic formula and see what you get :

a=12
b=3
c=0

thus
x = {-3 +/- [(3)^2 - 4(12)(0)]^(1/2)} / 2(12)
= {-3 +/- 3 } / 24
= 0 , -6/24
= 0, -1/4

factor out the 3x
3x(4x+1)=0 divide the 3x out
4x+1=0 one of the x's equals 0
4x=-1
x=-1/4

x=-1/4 and 0