2b+4=16/2b=

Solution for 2b+4=16/2b= equation:

2b+4=16/2b=

We move all terms to the left:

2b+4-(16/2b)=0


Domain of the equation: 2b)!=0

b!=0/1

b!=0

b∈R


We add all the numbers together, and all the variables

2b-(+16/2b)+4=0

We get rid of parentheses

2b-16/2b+4=0

We multiply all the terms by the denominator


2b*2b+4*2b-16=0

Wy multiply elements

4b^2+8b-16=0

a = 4; b = 8; c = -16;
Δ = b2-4ac
Δ = 82-4·4·(-16)
Δ = 320
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{320}=\sqrt{64*5}=\sqrt{64}*\sqrt{5}=8\sqrt{5}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-8\sqrt{5}}{2*4}=\frac{-8-8\sqrt{5}}{8}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+8\sqrt{5}}{2*4}=\frac{-8+8\sqrt{5}}{8}
$