2k-8=13/5k-12

Solution for 2k-8=13/5k-12 equation:

2k-8=13/5k-12

We move all terms to the left:

2k-8-(13/5k-12)=0


Domain of the equation: 5k-12)!=0

k∈R


We get rid of parentheses

2k-13/5k+12-8=0

We multiply all the terms by the denominator


2k*5k+12*5k-8*5k-13=0

Wy multiply elements

10k^2+60k-40k-13=0

We add all the numbers together, and all the variables

10k^2+20k-13=0

a = 10; b = 20; c = -13;
Δ = b2-4ac
Δ = 202-4·10·(-13)
Δ = 920
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:

$\sqrt{\Delta}=\sqrt{920}=\sqrt{4*230}=\sqrt{4}*\sqrt{230}=2\sqrt{230}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-2\sqrt{230}}{2*10}=\frac{-20-2\sqrt{230}}{20}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+2\sqrt{230}}{2*10}=\frac{-20+2\sqrt{230}}{20}
$