3x-12x=03x(x-4)=0

Solution for 3x-12x=03x(x-4)=0 equation:

3x-12x=03x(x-4)=0

We move all terms to the left:

3x-12x-(03x(x-4))=0

We add all the numbers together, and all the variables

-9x-(03x(x-4))=0


We calculate terms in parentheses: -(03x(x-4)), so:

03x(x-4)

We multiply parentheses

3x^2-12x

Back to the equation:

-(3x^2-12x)


We get rid of parentheses

-3x^2-9x+12x=0

We add all the numbers together, and all the variables

-3x^2+3x=0

a = -3; b = 3; c = 0;
Δ = b2-4ac
Δ = 32-4·(-3)·0
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3}{2*-3}=\frac{-6}{-6}
=1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3}{2*-3}=\frac{0}{-6}
=0
$