2k-8=8/5k-12

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

2k-8=8/5k-12

We move all terms to the left:

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


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

k∈R


We get rid of parentheses

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

We multiply all the terms by the denominator


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

Wy multiply elements

10k^2+60k-40k-8=0

We add all the numbers together, and all the variables

10k^2+20k-8=0

a = 10; b = 20; c = -8;
Δ = b2-4ac
Δ = 202-4·10·(-8)
Δ = 720
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:

$\sqrt{\Delta}=\sqrt{720}=\sqrt{144*5}=\sqrt{144}*\sqrt{5}=12\sqrt{5}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-12\sqrt{5}}{2*10}=\frac{-20-12\sqrt{5}}{20}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+12\sqrt{5}}{2*10}=\frac{-20+12\sqrt{5}}{20}
$