k+3/3=8/k-2

Solution for k+3/3=8/k-2 equation:

k+3/3=8/k-2

We move all terms to the left:

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


Domain of the equation: k-2)!=0

k∈R


We add all the numbers together, and all the variables

k-(8/k-2)+1=0

We get rid of parentheses

k-8/k+2+1=0

We multiply all the terms by the denominator


k*k+2*k+1*k-8=0

We add all the numbers together, and all the variables

3k+k*k-8=0

Wy multiply elements

k^2+3k-8=0

a = 1; b = 3; c = -8;
Δ = b2-4ac
Δ = 32-4·1·(-8)
Δ = 41
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-\sqrt{41}}{2*1}=\frac{-3-\sqrt{41}}{2}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{41}}{2*1}=\frac{-3+\sqrt{41}}{2}
$