(1)/(2)w-(3)/(4)=(2)/(3)w+2

Solution for (1)/(2)w-(3)/(4)=(2)/(3)w+2 equation:

(1)/(2)w-(3)/(4)=(2)/(3)w+2

We move all terms to the left:

(1)/(2)w-(3)/(4)-((2)/(3)w+2)=0


Domain of the equation: 2w!=0

w!=0/2

w!=0

w∈R



Domain of the equation: 3w+2)!=0

w∈R


We get rid of parentheses

1/2w-2/3w-2-3/4=0

We calculate fractions


(-54w^2)/96w^2+48w/96w^2+(-64w)/96w^2-2=0

We multiply all the terms by the denominator


(-54w^2)+48w+(-64w)-2*96w^2=0

Wy multiply elements

(-54w^2)-192w^2+48w+(-64w)=0

We get rid of parentheses

-54w^2-192w^2+48w-64w=0

We add all the numbers together, and all the variables

-246w^2-16w=0

a = -246; b = -16; c = 0;
Δ = b2-4ac
Δ = -162-4·(-246)·0
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-16}{2*-246}=\frac{0}{-492}
=0
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+16}{2*-246}=\frac{32}{-492}
=-8/123
$