2/k+4-3/k=7/5k

Solution for 2/k+4-3/k=7/5k equation:

2/k+4-3/k=7/5k

We move all terms to the left:

2/k+4-3/k-(7/5k)=0


Domain of the equation: k!=0

k∈R



Domain of the equation: 5k)!=0

k!=0/1

k!=0

k∈R


We add all the numbers together, and all the variables

2/k-3/k-(+7/5k)+4=0

We get rid of parentheses

2/k-3/k-7/5k+4=0

We calculate fractions


(-15k+2)/5k^2+(-7k)/5k^2+4=0

We multiply all the terms by the denominator


(-15k+2)+(-7k)+4*5k^2=0

Wy multiply elements

20k^2+(-15k+2)+(-7k)=0

We get rid of parentheses

20k^2-15k-7k+2=0

We add all the numbers together, and all the variables

20k^2-22k+2=0

a = 20; b = -22; c = +2;
Δ = b2-4ac
Δ = -222-4·20·2
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-18}{2*20}=\frac{4}{40}
=1/10
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+18}{2*20}=\frac{40}{40}
=1
$