1/2b+3+3/2b=11+2b

Solution for 1/2b+3+3/2b=11+2b equation:

1/2b+3+3/2b=11+2b

We move all terms to the left:

1/2b+3+3/2b-(11+2b)=0


Domain of the equation: 2b!=0

b!=0/2

b!=0

b∈R


We add all the numbers together, and all the variables

1/2b+3/2b-(2b+11)+3=0

We get rid of parentheses

1/2b+3/2b-2b-11+3=0

We multiply all the terms by the denominator


-2b*2b-11*2b+3*2b+1+3=0

We add all the numbers together, and all the variables

-2b*2b-11*2b+3*2b+4=0

Wy multiply elements

-4b^2-22b+6b+4=0

We add all the numbers together, and all the variables

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

a = -4; b = -16; c = +4;
Δ = b2-4ac
Δ = -162-4·(-4)·4
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{-(-16)-8\sqrt{5}}{2*-4}=\frac{16-8\sqrt{5}}{-8}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+8\sqrt{5}}{2*-4}=\frac{16+8\sqrt{5}}{-8}
$