2/7k-9/5=3-3/2k

Solution for 2/7k-9/5=3-3/2k equation:

2/7k-9/5=3-3/2k

We move all terms to the left:

2/7k-9/5-(3-3/2k)=0


Domain of the equation: 7k!=0

k!=0/7

k!=0

k∈R



Domain of the equation: 2k)!=0

k!=0/1

k!=0

k∈R


We add all the numbers together, and all the variables

2/7k-(-3/2k+3)-9/5=0

We get rid of parentheses

2/7k+3/2k-3-9/5=0

We calculate fractions


(-252k^2)/350k^2+100k/350k^2+525k/350k^2-3=0

We multiply all the terms by the denominator


(-252k^2)+100k+525k-3*350k^2=0

We add all the numbers together, and all the variables

(-252k^2)+625k-3*350k^2=0

Wy multiply elements

(-252k^2)-1050k^2+625k=0

We get rid of parentheses

-252k^2-1050k^2+625k=0

We add all the numbers together, and all the variables

-1302k^2+625k=0

a = -1302; b = 625; c = 0;
Δ = b2-4ac
Δ = 6252-4·(-1302)·0
Δ = 390625
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{390625}=625$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(625)-625}{2*-1302}=\frac{-1250}{-2604}
=625/1302
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(625)+625}{2*-1302}=\frac{0}{-2604}
=0
$