1/5x+5=191/2x

Solution for 1/5x+5=191/2x equation:

1/5x+5=191/2x

We move all terms to the left:

1/5x+5-(191/2x)=0


Domain of the equation: 5x!=0

x!=0/5

x!=0

x∈R



Domain of the equation: 2x)!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

1/5x-(+191/2x)+5=0

We get rid of parentheses

1/5x-191/2x+5=0

We calculate fractions


2x/10x^2+(-955x)/10x^2+5=0

We multiply all the terms by the denominator


2x+(-955x)+5*10x^2=0

Wy multiply elements

50x^2+2x+(-955x)=0

We get rid of parentheses

50x^2+2x-955x=0

We add all the numbers together, and all the variables

50x^2-953x=0

a = 50; b = -953; c = 0;
Δ = b2-4ac
Δ = -9532-4·50·0
Δ = 908209
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{908209}=953$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-953)-953}{2*50}=\frac{0}{100}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-953)+953}{2*50}=\frac{1906}{100}
=19+3/50
$