2-2/4k=1/8k+9

Solution for 2-2/4k=1/8k+9 equation:

2-2/4k=1/8k+9

We move all terms to the left:

2-2/4k-(1/8k+9)=0


Domain of the equation: 4k!=0

k!=0/4

k!=0

k∈R



Domain of the equation: 8k+9)!=0

k∈R


We get rid of parentheses

-2/4k-1/8k-9+2=0

We calculate fractions


(-16k)/32k^2+(-4k)/32k^2-9+2=0

We add all the numbers together, and all the variables

(-16k)/32k^2+(-4k)/32k^2-7=0

We multiply all the terms by the denominator


(-16k)+(-4k)-7*32k^2=0

Wy multiply elements

-224k^2+(-16k)+(-4k)=0

We get rid of parentheses

-224k^2-16k-4k=0

We add all the numbers together, and all the variables

-224k^2-20k=0

a = -224; b = -20; c = 0;
Δ = b2-4ac
Δ = -202-4·(-224)·0
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-20}{2*-224}=\frac{0}{-448}
=0
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+20}{2*-224}=\frac{40}{-448}
=-5/56
$