5/4b+7=7/8b+19

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

5/4b+7=7/8b+19

We move all terms to the left:

5/4b+7-(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+7=0

We calculate fractions


40b/32b^2+(-28b)/32b^2-19+7=0

We add all the numbers together, and all the variables

40b/32b^2+(-28b)/32b^2-12=0

We multiply all the terms by the denominator


40b+(-28b)-12*32b^2=0

Wy multiply elements

-384b^2+40b+(-28b)=0

We get rid of parentheses

-384b^2+40b-28b=0

We add all the numbers together, and all the variables

-384b^2+12b=0

a = -384; b = 12; c = 0;
Δ = b2-4ac
Δ = 122-4·(-384)·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*-384}=\frac{-24}{-768}
=1/32
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+12}{2*-384}=\frac{0}{-768}
=0
$