6+(4/5x)=9/10x

Solution for 6+(4/5x)=9/10x equation:

6+(4/5x)=9/10x

We move all terms to the left:

6+(4/5x)-(9/10x)=0


Domain of the equation: 5x)!=0

x!=0/1

x!=0

x∈R



Domain of the equation: 10x)!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

(+4/5x)-(+9/10x)+6=0

We get rid of parentheses

4/5x-9/10x+6=0

We calculate fractions


40x/50x^2+(-45x)/50x^2+6=0

We multiply all the terms by the denominator


40x+(-45x)+6*50x^2=0

Wy multiply elements

300x^2+40x+(-45x)=0

We get rid of parentheses

300x^2+40x-45x=0

We add all the numbers together, and all the variables

300x^2-5x=0

a = 300; b = -5; c = 0;
Δ = b2-4ac
Δ = -52-4·300·0
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-5}{2*300}=\frac{0}{600}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+5}{2*300}=\frac{10}{600}
=1/60
$