2(1/5x)-23=7(3/10x)+34

Solution for 2(1/5x)-23=7(3/10x)+34 equation:

2(1/5x)-23=7(3/10x)+34

We move all terms to the left:

2(1/5x)-23-(7(3/10x)+34)=0


Domain of the equation: 5x)!=0

x!=0/1

x!=0

x∈R



Domain of the equation: 10x)+34)!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

2(+1/5x)-(7(+3/10x)+34)-23=0

We multiply parentheses

2x-(7(+3/10x)+34)-23=0

We multiply all the terms by the denominator


2x*10x)+34)-(7(-23*10x)+34)+3=0

Wy multiply elements

20x^2-230x=0

a = 20; b = -230; c = 0;
Δ = b2-4ac
Δ = -2302-4·20·0
Δ = 52900
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{52900}=230$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-230)-230}{2*20}=\frac{0}{40}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-230)+230}{2*20}=\frac{460}{40}
=11+1/2
$