432x+342=964/5x

Solution for 432x+342=964/5x equation:

432x+342=964/5x

We move all terms to the left:

432x+342-(964/5x)=0


Domain of the equation: 5x)!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

432x-(+964/5x)+342=0

We get rid of parentheses

432x-964/5x+342=0

We multiply all the terms by the denominator


432x*5x+342*5x-964=0

Wy multiply elements

2160x^2+1710x-964=0

a = 2160; b = 1710; c = -964;
Δ = b2-4ac
Δ = 17102-4·2160·(-964)
Δ = 11253060
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:

$\sqrt{\Delta}=\sqrt{11253060}=\sqrt{36*312585}=\sqrt{36}*\sqrt{312585}=6\sqrt{312585}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1710)-6\sqrt{312585}}{2*2160}=\frac{-1710-6\sqrt{312585}}{4320}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1710)+6\sqrt{312585}}{2*2160}=\frac{-1710+6\sqrt{312585}}{4320}
$