4k-6-16/2k+3+2=0

Solution for 4k-6-16/2k+3+2=0 equation:

4k-6-16/2k+3+2=0


Domain of the equation: 2k!=0

k!=0/2

k!=0

k∈R


We add all the numbers together, and all the variables

4k-16/2k-1=0

We multiply all the terms by the denominator


4k*2k-1*2k-16=0

Wy multiply elements

8k^2-2k-16=0

a = 8; b = -2; c = -16;
Δ = b2-4ac
Δ = -22-4·8·(-16)
Δ = 516
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:

$\sqrt{\Delta}=\sqrt{516}=\sqrt{4*129}=\sqrt{4}*\sqrt{129}=2\sqrt{129}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{129}}{2*8}=\frac{2-2\sqrt{129}}{16}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{129}}{2*8}=\frac{2+2\sqrt{129}}{16}
$