2x+25=11/5x=20

Solution for 2x+25=11/5x=20 equation:

2x+25=11/5x=20

We move all terms to the left:

2x+25-(11/5x)=0


Domain of the equation: 5x)!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

2x-(+11/5x)+25=0

We get rid of parentheses

2x-11/5x+25=0

We multiply all the terms by the denominator


2x*5x+25*5x-11=0

Wy multiply elements

10x^2+125x-11=0

a = 10; b = 125; c = -11;
Δ = b2-4ac
Δ = 1252-4·10·(-11)
Δ = 16065
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:

$\sqrt{\Delta}=\sqrt{16065}=\sqrt{9*1785}=\sqrt{9}*\sqrt{1785}=3\sqrt{1785}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(125)-3\sqrt{1785}}{2*10}=\frac{-125-3\sqrt{1785}}{20}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(125)+3\sqrt{1785}}{2*10}=\frac{-125+3\sqrt{1785}}{20}
$