-2/9k+4/5=2-6/5k

Solution for -2/9k+4/5=2-6/5k equation:

-2/9k+4/5=2-6/5k

We move all terms to the left:

-2/9k+4/5-(2-6/5k)=0


Domain of the equation: 9k!=0

k!=0/9

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

-2/9k-(-6/5k+2)+4/5=0

We get rid of parentheses

-2/9k+6/5k-2+4/5=0

We calculate fractions


(-250k)/1125k^2+54k/1125k^2+36k/1125k^2-2=0

We multiply all the terms by the denominator


(-250k)+54k+36k-2*1125k^2=0

We add all the numbers together, and all the variables

90k+(-250k)-2*1125k^2=0

Wy multiply elements

-2250k^2+90k+(-250k)=0

We get rid of parentheses

-2250k^2+90k-250k=0

We add all the numbers together, and all the variables

-2250k^2-160k=0

a = -2250; b = -160; c = 0;
Δ = b2-4ac
Δ = -1602-4·(-2250)·0
Δ = 25600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25600}=160$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-160)-160}{2*-2250}=\frac{0}{-4500}
=0
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-160)+160}{2*-2250}=\frac{320}{-4500}
=-16/225
$