(2/3k)+3-2k=11

Solution for (2/3k)+3-2k=11 equation:

(2/3k)+3-2k=11

We move all terms to the left:

(2/3k)+3-2k-(11)=0


Domain of the equation: 3k)!=0

k!=0/1

k!=0

k∈R


We add all the numbers together, and all the variables

(+2/3k)-2k+3-11=0

We add all the numbers together, and all the variables

-2k+(+2/3k)-8=0

We get rid of parentheses

-2k+2/3k-8=0

We multiply all the terms by the denominator


-2k*3k-8*3k+2=0

Wy multiply elements

-6k^2-24k+2=0

a = -6; b = -24; c = +2;
Δ = b2-4ac
Δ = -242-4·(-6)·2
Δ = 624
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:

$\sqrt{\Delta}=\sqrt{624}=\sqrt{16*39}=\sqrt{16}*\sqrt{39}=4\sqrt{39}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-4\sqrt{39}}{2*-6}=\frac{24-4\sqrt{39}}{-12}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+4\sqrt{39}}{2*-6}=\frac{24+4\sqrt{39}}{-12}
$