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

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

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

We move all terms to the left:

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


Domain of the equation: (4-x)!=0

We move all terms containing x to the left, all other terms to the right

-x!=-4

x!=-4/-1

x!=+4

x∈R


We add all the numbers together, and all the variables

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

We multiply parentheses ..

(-3x^2+12x-2x+8)/(-1x+4)-2=0

We multiply all the terms by the denominator


(-3x^2+12x-2x+8)-2*(-1x+4)=0

We multiply parentheses

(-3x^2+12x-2x+8)+2x-8=0

We get rid of parentheses

-3x^2+12x-2x+2x+8-8=0

We add all the numbers together, and all the variables

-3x^2+12x=0

a = -3; b = 12; c = 0;
Δ = b2-4ac
Δ = 122-4·(-3)·0
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-12}{2*-3}=\frac{-24}{-6}
=+4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+12}{2*-3}=\frac{0}{-6}
=0
$