3x+4=3x(x+12)

Solution for 3x+4=3x(x+12) equation:

3x+4=3x(x+12)

We move all terms to the left:

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


We calculate terms in parentheses: -(3x(x+12)), so:

3x(x+12)

We multiply parentheses

3x^2+36x

Back to the equation:

-(3x^2+36x)


We get rid of parentheses

-3x^2+3x-36x+4=0

We add all the numbers together, and all the variables

-3x^2-33x+4=0

a = -3; b = -33; c = +4;
Δ = b2-4ac
Δ = -332-4·(-3)·4
Δ = 1137
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-33)-\sqrt{1137}}{2*-3}=\frac{33-\sqrt{1137}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-33)+\sqrt{1137}}{2*-3}=\frac{33+\sqrt{1137}}{-6}
$