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

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

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

We move all terms to the left:

2/4c+4-1/5c-(-9)=0


Domain of the equation: 4c!=0

c!=0/4

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

2/4c-1/5c+13=0

We calculate fractions


10c/20c^2+(-4c)/20c^2+13=0

We multiply all the terms by the denominator


10c+(-4c)+13*20c^2=0

Wy multiply elements

260c^2+10c+(-4c)=0

We get rid of parentheses

260c^2+10c-4c=0

We add all the numbers together, and all the variables

260c^2+6c=0

a = 260; b = 6; c = 0;
Δ = b2-4ac
Δ = 62-4·260·0
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-6}{2*260}=\frac{-12}{520}
=-3/130
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6}{2*260}=\frac{0}{520}
=0
$