3x+1=2-(x-4)3x

Solution for 3x+1=2-(x-4)3x equation:

3x+1=2-(x-4)3x

We move all terms to the left:

3x+1-(2-(x-4)3x)=0


We calculate terms in parentheses: -(2-(x-4)3x), so:

2-(x-4)3x

determiningTheFunctionDomain
-(x-4)3x+2

We multiply parentheses

-3x^2+12x+2

Back to the equation:

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


We get rid of parentheses

3x^2-12x+3x-2+1=0

We add all the numbers together, and all the variables

3x^2-9x-1=0

a = 3; b = -9; c = -1;
Δ = b2-4ac
Δ = -92-4·3·(-1)
Δ = 93
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{-(-9)-\sqrt{93}}{2*3}=\frac{9-\sqrt{93}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+\sqrt{93}}{2*3}=\frac{9+\sqrt{93}}{6}
$