-3/w+3=5/4w+12+1

Solution for -3/w+3=5/4w+12+1 equation:

-3/w+3=5/4w+12+1

We move all terms to the left:

-3/w+3-(5/4w+12+1)=0


Domain of the equation: w!=0

w∈R



Domain of the equation: 4w+12+1)!=0

We move all terms containing w to the left, all other terms to the right

4w+1)!=-12

w∈R


We add all the numbers together, and all the variables

-3/w-(5/4w+13)+3=0

We get rid of parentheses

-3/w-5/4w-13+3=0

We calculate fractions


(-12w)/4w^2+(-5w)/4w^2-13+3=0

We add all the numbers together, and all the variables

(-12w)/4w^2+(-5w)/4w^2-10=0

We multiply all the terms by the denominator


(-12w)+(-5w)-10*4w^2=0

Wy multiply elements

-40w^2+(-12w)+(-5w)=0

We get rid of parentheses

-40w^2-12w-5w=0

We add all the numbers together, and all the variables

-40w^2-17w=0

a = -40; b = -17; c = 0;
Δ = b2-4ac
Δ = -172-4·(-40)·0
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-17)-17}{2*-40}=\frac{0}{-80}
=0
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-17)+17}{2*-40}=\frac{34}{-80}
=-17/40
$