X=3,4x+2006-(4(3+1)+x+(16/3)=

Solution for X=3,4x+2006-(4(3+1)+x+(16/3)= equation:

X=3.4X+2006-(4(3+1)+X+(16/3)=

We move all terms to the left:

X-(3.4X+2006-(4(3+1)+X+(16/3))=0

We add all the numbers together, and all the variables

X-(3.4X+2006-(44+X+(+16/3))=0

We multiply all the terms by the denominator


X*3))-(3.4X+2006-(44+X+(+16=0

We add all the numbers together, and all the variables

X+X*3))-(3.4X+2006-(44=0

Wy multiply elements

3X^2+X=0

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

$\sqrt{\Delta}=\sqrt{1}=1$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-1}{2*3}=\frac{-2}{6}
=-1/3
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+1}{2*3}=\frac{0}{6}
=0
$