2-3/4k+9=1/8k+9

Solution for 2-3/4k+9=1/8k+9 equation:

2-3/4k+9=1/8k+9

We move all terms to the left:

2-3/4k+9-(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 add all the numbers together, and all the variables

-3/4k-(1/8k+9)+11=0

We get rid of parentheses

-3/4k-1/8k-9+11=0

We calculate fractions


(-24k)/32k^2+(-4k)/32k^2-9+11=0

We add all the numbers together, and all the variables

(-24k)/32k^2+(-4k)/32k^2+2=0

We multiply all the terms by the denominator


(-24k)+(-4k)+2*32k^2=0

Wy multiply elements

64k^2+(-24k)+(-4k)=0

We get rid of parentheses

64k^2-24k-4k=0

We add all the numbers together, and all the variables

64k^2-28k=0

a = 64; b = -28; c = 0;
Δ = b2-4ac
Δ = -282-4·64·0
Δ = 784
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{784}=28$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-28}{2*64}=\frac{0}{128}
=0
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+28}{2*64}=\frac{56}{128}
=7/16
$