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

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

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

We move all terms to the left:

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


Domain of the equation: 2b)!=0

b!=0/1

b!=0

b∈R


We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We multiply all the terms by the denominator


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

Wy multiply elements

6b^2+22b-6b-1=0

We add all the numbers together, and all the variables

6b^2+16b-1=0

a = 6; b = 16; c = -1;
Δ = b2-4ac
Δ = 162-4·6·(-1)
Δ = 280
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:

$\sqrt{\Delta}=\sqrt{280}=\sqrt{4*70}=\sqrt{4}*\sqrt{70}=2\sqrt{70}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-2\sqrt{70}}{2*6}=\frac{-16-2\sqrt{70}}{12}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+2\sqrt{70}}{2*6}=\frac{-16+2\sqrt{70}}{12}
$