2k-1=3/4k+9

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

2k-1=3/4k+9

We move all terms to the left:

2k-1-(3/4k+9)=0


Domain of the equation: 4k+9)!=0

k∈R


We get rid of parentheses

2k-3/4k-9-1=0

We multiply all the terms by the denominator


2k*4k-9*4k-1*4k-3=0

Wy multiply elements

8k^2-36k-4k-3=0

We add all the numbers together, and all the variables

8k^2-40k-3=0

a = 8; b = -40; c = -3;
Δ = b2-4ac
Δ = -402-4·8·(-3)
Δ = 1696
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1696}=\sqrt{16*106}=\sqrt{16}*\sqrt{106}=4\sqrt{106}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-4\sqrt{106}}{2*8}=\frac{40-4\sqrt{106}}{16}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+4\sqrt{106}}{2*8}=\frac{40+4\sqrt{106}}{16}
$