1/4b+1/2b=3

Solution for 1/4b+1/2b=3 equation:

1/4b+1/2b=3

We move all terms to the left:

1/4b+1/2b-(3)=0


Domain of the equation: 4b!=0

b!=0/4

b!=0

b∈R



Domain of the equation: 2b!=0

b!=0/2

b!=0

b∈R


We calculate fractions


2b/8b^2+4b/8b^2-3=0

We multiply all the terms by the denominator


2b+4b-3*8b^2=0

We add all the numbers together, and all the variables

6b-3*8b^2=0

Wy multiply elements

-24b^2+6b=0

a = -24; b = 6; c = 0;
Δ = b2-4ac
Δ = 62-4·(-24)·0
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-6}{2*-24}=\frac{-12}{-48}
=1/4
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6}{2*-24}=\frac{0}{-48}
=0
$