42/3x+33/4=-71/2x-5

Solution for 42/3x+33/4=-71/2x-5 equation:

42/3x+33/4=-71/2x-5

We move all terms to the left:

42/3x+33/4-(-71/2x-5)=0


Domain of the equation: 3x!=0

x!=0/3

x!=0

x∈R



Domain of the equation: 2x-5)!=0

x∈R


We get rid of parentheses

42/3x+71/2x+5+33/4=0

We calculate fractions


396x^2/96x^2+1344x/96x^2+3408x/96x^2+5=0

We multiply all the terms by the denominator


396x^2+1344x+3408x+5*96x^2=0

We add all the numbers together, and all the variables

396x^2+4752x+5*96x^2=0

Wy multiply elements

396x^2+480x^2+4752x=0

We add all the numbers together, and all the variables

876x^2+4752x=0

a = 876; b = 4752; c = 0;
Δ = b2-4ac
Δ = 47522-4·876·0
Δ = 22581504
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{22581504}=4752$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4752)-4752}{2*876}=\frac{-9504}{1752}
=-5+31/73
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4752)+4752}{2*876}=\frac{0}{1752}
=0
$