x^2-4x=32/3-2x-x^2

Solution for x^2-4x=32/3-2x-x^2 equation:

x^2-4x=32/3-2x-x^2

We move all terms to the left:

x^2-4x-(32/3-2x-x^2)=0


Domain of the equation: 3-2x-x^2)!=0

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

-x^2)-2x!=-3

x∈R


We get rid of parentheses

x^2+x^2+2x-4x-32/3=0

We multiply all the terms by the denominator


x^2*3+x^2*3+2x*3-4x*3-32=0

Wy multiply elements

3x^2+3x^2+6x-12x-32=0

We add all the numbers together, and all the variables

6x^2-6x-32=0

a = 6; b = -6; c = -32;
Δ = b2-4ac
Δ = -62-4·6·(-32)
Δ = 804
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{804}=\sqrt{4*201}=\sqrt{4}*\sqrt{201}=2\sqrt{201}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{201}}{2*6}=\frac{6-2\sqrt{201}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{201}}{2*6}=\frac{6+2\sqrt{201}}{12}
$