3/5c-25=5+c

Solution for 3/5c-25=5+c equation:

3/5c-25=5+c

We move all terms to the left:

3/5c-25-(5+c)=0


Domain of the equation: 5c!=0

c!=0/5

c!=0

c∈R


We add all the numbers together, and all the variables

3/5c-(c+5)-25=0

We get rid of parentheses

3/5c-c-5-25=0

We multiply all the terms by the denominator


-c*5c-5*5c-25*5c+3=0

Wy multiply elements

-5c^2-25c-125c+3=0

We add all the numbers together, and all the variables

-5c^2-150c+3=0

a = -5; b = -150; c = +3;
Δ = b2-4ac
Δ = -1502-4·(-5)·3
Δ = 22560
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:

$\sqrt{\Delta}=\sqrt{22560}=\sqrt{16*1410}=\sqrt{16}*\sqrt{1410}=4\sqrt{1410}$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-150)-4\sqrt{1410}}{2*-5}=\frac{150-4\sqrt{1410}}{-10}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-150)+4\sqrt{1410}}{2*-5}=\frac{150+4\sqrt{1410}}{-10}
$