5x/10+20x/5x=40

Solution for 5x/10+20x/5x=40 equation:

5x/10+20x/5x=40

We move all terms to the left:

5x/10+20x/5x-(40)=0


Domain of the equation: 5x!=0

x!=0/5

x!=0

x∈R


We calculate fractions


25x^2/50x+200x/50x-40=0

We multiply all the terms by the denominator


25x^2+200x-40*50x=0

Wy multiply elements

25x^2+200x-2000x=0

We add all the numbers together, and all the variables

25x^2-1800x=0

a = 25; b = -1800; c = 0;
Δ = b2-4ac
Δ = -18002-4·25·0
Δ = 3240000
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3240000}=1800$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1800)-1800}{2*25}=\frac{0}{50}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1800)+1800}{2*25}=\frac{3600}{50}
=72
$