0.3+(x/10)=(1/2x)-0.4

Solution for 0.3+(x/10)=(1/2x)-0.4 equation:

0.3+(x/10)=(1/2x)-0.4

We move all terms to the left:

0.3+(x/10)-((1/2x)-0.4)=0


Domain of the equation: 2x)-0.4)!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

(+x/10)-((+1/2x)-0.4)+0.3=0

We get rid of parentheses

x/10-((+1/2x)-0.4)+0.3=0

We calculate fractions


2x^2/20x+()/20x+0.3=0

We multiply all the terms by the denominator


2x^2+(0.3)*20x+()=0

We add all the numbers together, and all the variables

2x^2+(0.3)*20x=0

We multiply parentheses

2x^2+6x=0

a = 2; b = 6; c = 0;
Δ = b2-4ac
Δ = 62-4·2·0
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-6}{2*2}=\frac{-12}{4}
=-3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6}{2*2}=\frac{0}{4}
=0
$