5(2k-3)-3(k+4)/3k+2;k=-2

Solution for 5(2k-3)-3(k+4)/3k+2;k=-2 equation:

5(2k-3)-3(k+4)/3k+2k=-2

We move all terms to the left:

5(2k-3)-3(k+4)/3k+2k-(-2)=0


Domain of the equation: 3k!=0

k!=0/3

k!=0

k∈R


We add all the numbers together, and all the variables

2k+5(2k-3)-3(k+4)/3k+2=0

We multiply parentheses

2k+10k-3(k+4)/3k-15+2=0

We multiply all the terms by the denominator


2k*3k+10k*3k-3(k+4)-15*3k+2*3k=0

We multiply parentheses

2k*3k+10k*3k-3k-15*3k+2*3k-12=0

Wy multiply elements

6k^2+30k^2-3k-45k+6k-12=0

We add all the numbers together, and all the variables

36k^2-42k-12=0

a = 36; b = -42; c = -12;
Δ = b2-4ac
Δ = -422-4·36·(-12)
Δ = 3492
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:

$\sqrt{\Delta}=\sqrt{3492}=\sqrt{36*97}=\sqrt{36}*\sqrt{97}=6\sqrt{97}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-42)-6\sqrt{97}}{2*36}=\frac{42-6\sqrt{97}}{72}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-42)+6\sqrt{97}}{2*36}=\frac{42+6\sqrt{97}}{72}
$