3x^2=(x-4)+12x

Solution for 3x^2=(x-4)+12x equation:

3x^2=(x-4)+12x

We move all terms to the left:

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


We calculate terms in parentheses: -((x-4)+12x), so:

(x-4)+12x

We add all the numbers together, and all the variables

12x+(x-4)

We get rid of parentheses

12x+x-4

We add all the numbers together, and all the variables

13x-4

Back to the equation:

-(13x-4)


We get rid of parentheses

3x^2-13x+4=0

a = 3; b = -13; c = +4;
Δ = b2-4ac
Δ = -132-4·3·4
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-11}{2*3}=\frac{2}{6}
=1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+11}{2*3}=\frac{24}{6}
=4
$