3/2b+8+1/2b=12+2b

Solution for 3/2b+8+1/2b=12+2b equation:

3/2b+8+1/2b=12+2b

We move all terms to the left:

3/2b+8+1/2b-(12+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

3/2b+1/2b-(2b+12)+8=0

We get rid of parentheses

3/2b+1/2b-2b-12+8=0

We multiply all the terms by the denominator


-2b*2b-12*2b+8*2b+3+1=0

We add all the numbers together, and all the variables

-2b*2b-12*2b+8*2b+4=0

Wy multiply elements

-4b^2-24b+16b+4=0

We add all the numbers together, and all the variables

-4b^2-8b+4=0

a = -4; b = -8; c = +4;
Δ = b2-4ac
Δ = -82-4·(-4)·4
Δ = 128
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:

$\sqrt{\Delta}=\sqrt{128}=\sqrt{64*2}=\sqrt{64}*\sqrt{2}=8\sqrt{2}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-8\sqrt{2}}{2*-4}=\frac{8-8\sqrt{2}}{-8}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+8\sqrt{2}}{2*-4}=\frac{8+8\sqrt{2}}{-8}
$