2/5c-20=-2c+4

Solution for 2/5c-20=-2c+4 equation:

2/5c-20=-2c+4

We move all terms to the left:

2/5c-20-(-2c+4)=0


Domain of the equation: 5c!=0

c!=0/5

c!=0

c∈R


We get rid of parentheses

2/5c+2c-4-20=0

We multiply all the terms by the denominator


2c*5c-4*5c-20*5c+2=0

Wy multiply elements

10c^2-20c-100c+2=0

We add all the numbers together, and all the variables

10c^2-120c+2=0

a = 10; b = -120; c = +2;
Δ = b2-4ac
Δ = -1202-4·10·2
Δ = 14320
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{14320}=\sqrt{16*895}=\sqrt{16}*\sqrt{895}=4\sqrt{895}$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-120)-4\sqrt{895}}{2*10}=\frac{120-4\sqrt{895}}{20}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-120)+4\sqrt{895}}{2*10}=\frac{120+4\sqrt{895}}{20}
$