5/4b=7/8b+19

Solution for 5/4b=7/8b+19 equation:

5/4b=7/8b+19

We move all terms to the left:

5/4b-(7/8b+19)=0


Domain of the equation: 4b!=0

b!=0/4

b!=0

b∈R



Domain of the equation: 8b+19)!=0

b∈R


We get rid of parentheses

5/4b-7/8b-19=0

We calculate fractions


40b/32b^2+(-28b)/32b^2-19=0

We multiply all the terms by the denominator


40b+(-28b)-19*32b^2=0

Wy multiply elements

-608b^2+40b+(-28b)=0

We get rid of parentheses

-608b^2+40b-28b=0

We add all the numbers together, and all the variables

-608b^2+12b=0

a = -608; b = 12; c = 0;
Δ = b2-4ac
Δ = 122-4·(-608)·0
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-12}{2*-608}=\frac{-24}{-1216}
=3/152
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+12}{2*-608}=\frac{0}{-1216}
=0
$