(7)/(8)b-32=(3)/(4)b+22

Solution for (7)/(8)b-32=(3)/(4)b+22 equation:

(7)/(8)b-32=(3)/(4)b+22

We move all terms to the left:

(7)/(8)b-32-((3)/(4)b+22)=0


Domain of the equation: 8b!=0

b!=0/8

b!=0

b∈R



Domain of the equation: 4b+22)!=0

b∈R


We get rid of parentheses

7/8b-3/4b-22-32=0

We calculate fractions


28b/32b^2+(-24b)/32b^2-22-32=0

We add all the numbers together, and all the variables

28b/32b^2+(-24b)/32b^2-54=0

We multiply all the terms by the denominator


28b+(-24b)-54*32b^2=0

Wy multiply elements

-1728b^2+28b+(-24b)=0

We get rid of parentheses

-1728b^2+28b-24b=0

We add all the numbers together, and all the variables

-1728b^2+4b=0

a = -1728; b = 4; c = 0;
Δ = b2-4ac
Δ = 42-4·(-1728)·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*-1728}=\frac{-8}{-3456}
=1/432
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4}{2*-1728}=\frac{0}{-3456}
=0
$