3k+3=33/2k

Solution for 3k+3=33/2k equation:

3k+3=33/2k

We move all terms to the left:

3k+3-(33/2k)=0


Domain of the equation: 2k)!=0

k!=0/1

k!=0

k∈R


We add all the numbers together, and all the variables

3k-(+33/2k)+3=0

We get rid of parentheses

3k-33/2k+3=0

We multiply all the terms by the denominator


3k*2k+3*2k-33=0

Wy multiply elements

6k^2+6k-33=0

a = 6; b = 6; c = -33;
Δ = b2-4ac
Δ = 62-4·6·(-33)
Δ = 828
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{828}=\sqrt{36*23}=\sqrt{36}*\sqrt{23}=6\sqrt{23}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-6\sqrt{23}}{2*6}=\frac{-6-6\sqrt{23}}{12}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6\sqrt{23}}{2*6}=\frac{-6+6\sqrt{23}}{12}
$