40-3x/5x=6x+7/8

Solution for 40-3x/5x=6x+7/8 equation:

40-3x/5x=6x+7/8

We move all terms to the left:

40-3x/5x-(6x+7/8)=0


Domain of the equation: 5x!=0

x!=0/5

x!=0

x∈R


We add all the numbers together, and all the variables

-3x/5x-(+6x+7/8)+40=0

We get rid of parentheses

-3x/5x-6x+40-7/8=0

We calculate fractions


-6x+(-24x)/40x+(-35x)/40x+40=0

We multiply all the terms by the denominator


-6x*40x+(-24x)+(-35x)+40*40x=0

Wy multiply elements

-240x^2+(-24x)+(-35x)+1600x=0

We get rid of parentheses

-240x^2-24x-35x+1600x=0

We add all the numbers together, and all the variables

-240x^2+1541x=0

a = -240; b = 1541; c = 0;
Δ = b2-4ac
Δ = 15412-4·(-240)·0
Δ = 2374681
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2374681}=1541$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1541)-1541}{2*-240}=\frac{-3082}{-480}
=6+101/240
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1541)+1541}{2*-240}=\frac{0}{-480}
=0
$