7/3c+4/5c=6

Solution for 7/3c+4/5c=6 equation:

7/3c+4/5c=6

We move all terms to the left:

7/3c+4/5c-(6)=0


Domain of the equation: 3c!=0

c!=0/3

c!=0

c∈R



Domain of the equation: 5c!=0

c!=0/5

c!=0

c∈R


We calculate fractions


35c/15c^2+12c/15c^2-6=0

We multiply all the terms by the denominator


35c+12c-6*15c^2=0

We add all the numbers together, and all the variables

47c-6*15c^2=0

Wy multiply elements

-90c^2+47c=0

a = -90; b = 47; c = 0;
Δ = b2-4ac
Δ = 472-4·(-90)·0
Δ = 2209
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2209}=47$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(47)-47}{2*-90}=\frac{-94}{-180}
=47/90
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(47)+47}{2*-90}=\frac{0}{-180}
=0
$