2/3d+1=16-d=

Solution for 2/3d+1=16-d= equation:

2/3d+1=16-d=

We move all terms to the left:

2/3d+1-(16-d)=0


Domain of the equation: 3d!=0

d!=0/3

d!=0

d∈R


We add all the numbers together, and all the variables

2/3d-(-1d+16)+1=0

We get rid of parentheses

2/3d+1d-16+1=0

We multiply all the terms by the denominator


1d*3d-16*3d+1*3d+2=0

Wy multiply elements

3d^2-48d+3d+2=0

We add all the numbers together, and all the variables

3d^2-45d+2=0

a = 3; b = -45; c = +2;
Δ = b2-4ac
Δ = -452-4·3·2
Δ = 2001
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-45)-\sqrt{2001}}{2*3}=\frac{45-\sqrt{2001}}{6}
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-45)+\sqrt{2001}}{2*3}=\frac{45+\sqrt{2001}}{6}
$