1/2b+2=3b+12

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

1/2b+2=3b+12

We move all terms to the left:

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


Domain of the equation: 2b!=0

b!=0/2

b!=0

b∈R


We get rid of parentheses

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

We multiply all the terms by the denominator


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

Wy multiply elements

-6b^2-24b+4b+1=0

We add all the numbers together, and all the variables

-6b^2-20b+1=0

a = -6; b = -20; c = +1;
Δ = b2-4ac
Δ = -202-4·(-6)·1
Δ = 424
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:

$\sqrt{\Delta}=\sqrt{424}=\sqrt{4*106}=\sqrt{4}*\sqrt{106}=2\sqrt{106}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-2\sqrt{106}}{2*-6}=\frac{20-2\sqrt{106}}{-12}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+2\sqrt{106}}{2*-6}=\frac{20+2\sqrt{106}}{-12}
$