3/4b-3=7/8b+2

Solution for 3/4b-3=7/8b+2 equation:

3/4b-3=7/8b+2

We move all terms to the left:

3/4b-3-(7/8b+2)=0


Domain of the equation: 4b!=0

b!=0/4

b!=0

b∈R



Domain of the equation: 8b+2)!=0

b∈R


We get rid of parentheses

3/4b-7/8b-2-3=0

We calculate fractions


24b/32b^2+(-28b)/32b^2-2-3=0

We add all the numbers together, and all the variables

24b/32b^2+(-28b)/32b^2-5=0

We multiply all the terms by the denominator


24b+(-28b)-5*32b^2=0

Wy multiply elements

-160b^2+24b+(-28b)=0

We get rid of parentheses

-160b^2+24b-28b=0

We add all the numbers together, and all the variables

-160b^2-4b=0

a = -160; b = -4; c = 0;
Δ = b2-4ac
Δ = -42-4·(-160)·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4}{2*-160}=\frac{0}{-320}
=0
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4}{2*-160}=\frac{8}{-320}
=-1/40
$