5/4c+7=7/9c+13

Solution for 5/4c+7=7/9c+13 equation:

5/4c+7=7/9c+13

We move all terms to the left:

5/4c+7-(7/9c+13)=0


Domain of the equation: 4c!=0

c!=0/4

c!=0

c∈R



Domain of the equation: 9c+13)!=0

c∈R


We get rid of parentheses

5/4c-7/9c-13+7=0

We calculate fractions


45c/36c^2+(-28c)/36c^2-13+7=0

We add all the numbers together, and all the variables

45c/36c^2+(-28c)/36c^2-6=0

We multiply all the terms by the denominator


45c+(-28c)-6*36c^2=0

Wy multiply elements

-216c^2+45c+(-28c)=0

We get rid of parentheses

-216c^2+45c-28c=0

We add all the numbers together, and all the variables

-216c^2+17c=0

a = -216; b = 17; c = 0;
Δ = b2-4ac
Δ = 172-4·(-216)·0
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-17}{2*-216}=\frac{-34}{-432}
=17/216
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+17}{2*-216}=\frac{0}{-432}
=0
$