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

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

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

We move all terms to the left:

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


Domain of the equation: 4c!=0

c!=0/4

c!=0

c∈R



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

c∈R


We get rid of parentheses

5/4c-7/8c-19+7=0

We calculate fractions


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

We add all the numbers together, and all the variables

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

We multiply all the terms by the denominator


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

Wy multiply elements

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-384c^2+12c=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:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-12}{2*-384}=\frac{-24}{-768}
=1/32
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+12}{2*-384}=\frac{0}{-768}
=0
$