39/5x=4+1/2x

Solution for 39/5x=4+1/2x equation:

39/5x=4+1/2x

We move all terms to the left:

39/5x-(4+1/2x)=0


Domain of the equation: 5x!=0

x!=0/5

x!=0

x∈R



Domain of the equation: 2x)!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

39/5x-(1/2x+4)=0

We get rid of parentheses

39/5x-1/2x-4=0

We calculate fractions


78x/10x^2+(-5x)/10x^2-4=0

We multiply all the terms by the denominator


78x+(-5x)-4*10x^2=0

Wy multiply elements

-40x^2+78x+(-5x)=0

We get rid of parentheses

-40x^2+78x-5x=0

We add all the numbers together, and all the variables

-40x^2+73x=0

a = -40; b = 73; c = 0;
Δ = b2-4ac
Δ = 732-4·(-40)·0
Δ = 5329
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{5329}=73$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(73)-73}{2*-40}=\frac{-146}{-80}
=1+33/40
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(73)+73}{2*-40}=\frac{0}{-80}
=0
$