1+11/2b=3b+2

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

1+11/2b=3b+2

We move all terms to the left:

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


Domain of the equation: 2b!=0

b!=0/2

b!=0

b∈R


We get rid of parentheses

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

We multiply all the terms by the denominator


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

Wy multiply elements

-6b^2-4b+2b+11=0

We add all the numbers together, and all the variables

-6b^2-2b+11=0

a = -6; b = -2; c = +11;
Δ = b2-4ac
Δ = -22-4·(-6)·11
Δ = 268
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:

$\sqrt{\Delta}=\sqrt{268}=\sqrt{4*67}=\sqrt{4}*\sqrt{67}=2\sqrt{67}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{67}}{2*-6}=\frac{2-2\sqrt{67}}{-12}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{67}}{2*-6}=\frac{2+2\sqrt{67}}{-12}
$