2/3c+4-1/5c=-9

Solution for 2/3c+4-1/5c=-9 equation:

2/3c+4-1/5c=-9

We move all terms to the left:

2/3c+4-1/5c-(-9)=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 add all the numbers together, and all the variables

2/3c-1/5c+13=0

We calculate fractions


10c/15c^2+(-3c)/15c^2+13=0

We multiply all the terms by the denominator


10c+(-3c)+13*15c^2=0

Wy multiply elements

195c^2+10c+(-3c)=0

We get rid of parentheses

195c^2+10c-3c=0

We add all the numbers together, and all the variables

195c^2+7c=0

a = 195; b = 7; c = 0;
Δ = b2-4ac
Δ = 72-4·195·0
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-7}{2*195}=\frac{-14}{390}
=-7/195
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+7}{2*195}=\frac{0}{390}
=0
$