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

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

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

We move all terms to the left:

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


We calculate terms in parentheses: -(2(x-4)+12x), so:

2(x-4)+12x

We add all the numbers together, and all the variables

12x+2(x-4)

We multiply parentheses

12x+2x-8

We add all the numbers together, and all the variables

14x-8

Back to the equation:

-(14x-8)


We get rid of parentheses

3x^2-14x+8=0

a = 3; b = -14; c = +8;
Δ = b2-4ac
Δ = -142-4·3·8
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-10}{2*3}=\frac{4}{6}
=2/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+10}{2*3}=\frac{24}{6}
=4
$