3/2b+3+1/2b=10+2b

Solution for 3/2b+3+1/2b=10+2b equation:

3/2b+3+1/2b=10+2b

We move all terms to the left:

3/2b+3+1/2b-(10+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+10)+3=0

We get rid of parentheses

3/2b+1/2b-2b-10+3=0

We multiply all the terms by the denominator


-2b*2b-10*2b+3*2b+3+1=0

We add all the numbers together, and all the variables

-2b*2b-10*2b+3*2b+4=0

Wy multiply elements

-4b^2-20b+6b+4=0

We add all the numbers together, and all the variables

-4b^2-14b+4=0

a = -4; b = -14; c = +4;
Δ = b2-4ac
Δ = -142-4·(-4)·4
Δ = 260
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:

$\sqrt{\Delta}=\sqrt{260}=\sqrt{4*65}=\sqrt{4}*\sqrt{65}=2\sqrt{65}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-2\sqrt{65}}{2*-4}=\frac{14-2\sqrt{65}}{-8}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+2\sqrt{65}}{2*-4}=\frac{14+2\sqrt{65}}{-8}
$