3.50c+100=1/50c

Solution for 3.50c+100=1/50c equation:

3.50c+100=1/50c

We move all terms to the left:

3.50c+100-(1/50c)=0


Domain of the equation: 50c)!=0

c!=0/1

c!=0

c∈R


We add all the numbers together, and all the variables

3.50c-(+1/50c)+100=0

We get rid of parentheses

3.50c-1/50c+100=0

We multiply all the terms by the denominator


(3.50c)*50c+100*50c-1=0

We add all the numbers together, and all the variables

(+3.50c)*50c+100*50c-1=0

We multiply parentheses

150c^2+100*50c-1=0

Wy multiply elements

150c^2+5000c-1=0

a = 150; b = 5000; c = -1;
Δ = b2-4ac
Δ = 50002-4·150·(-1)
Δ = 25000600
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:

$\sqrt{\Delta}=\sqrt{25000600}=\sqrt{100*250006}=\sqrt{100}*\sqrt{250006}=10\sqrt{250006}$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5000)-10\sqrt{250006}}{2*150}=\frac{-5000-10\sqrt{250006}}{300}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5000)+10\sqrt{250006}}{2*150}=\frac{-5000+10\sqrt{250006}}{300}
$