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

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

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

We move all terms to the left:

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


Domain of the equation: 3b!=0

b!=0/3

b!=0

b∈R



Domain of the equation: 2b!=0

b!=0/2

b!=0

b∈R


We get rid of parentheses

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

We calculate fractions


8b/24b^2+3b/24b^2+(-3b)/24b^2-12=0

We multiply all the terms by the denominator


8b+3b+(-3b)-12*24b^2=0

We add all the numbers together, and all the variables

11b+(-3b)-12*24b^2=0

Wy multiply elements

-288b^2+11b+(-3b)=0

We get rid of parentheses

-288b^2+11b-3b=0

We add all the numbers together, and all the variables

-288b^2+8b=0

a = -288; b = 8; c = 0;
Δ = b2-4ac
Δ = 82-4·(-288)·0
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-8}{2*-288}=\frac{-16}{-576}
=1/36
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+8}{2*-288}=\frac{0}{-576}
=0
$