3/100x+19.40=x

Solution for 3/100x+19.40=x equation:

3/100x+19.40=x

We move all terms to the left:

3/100x+19.40-(x)=0


Domain of the equation: 100x!=0

x!=0/100

x!=0

x∈R


We add all the numbers together, and all the variables

-1x+3/100x+19.4=0

We multiply all the terms by the denominator


-1x*100x+(19.4)*100x+3=0

We multiply parentheses

-1x*100x+1940x+3=0

Wy multiply elements

-100x^2+1940x+3=0

a = -100; b = 1940; c = +3;
Δ = b2-4ac
Δ = 19402-4·(-100)·3
Δ = 3764800
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{3764800}=\sqrt{1600*2353}=\sqrt{1600}*\sqrt{2353}=40\sqrt{2353}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1940)-40\sqrt{2353}}{2*-100}=\frac{-1940-40\sqrt{2353}}{-200}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1940)+40\sqrt{2353}}{2*-100}=\frac{-1940+40\sqrt{2353}}{-200}
$