-8/u-3=-3/4u-12+1

Solution for -8/u-3=-3/4u-12+1 equation:

-8/u-3=-3/4u-12+1

We move all terms to the left:

-8/u-3-(-3/4u-12+1)=0


Domain of the equation: u!=0

u∈R



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

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

4u+1)!=12

u∈R


We add all the numbers together, and all the variables

-8/u-(-3/4u-11)-3=0

We get rid of parentheses

-8/u+3/4u+11-3=0

We calculate fractions


(-32u)/4u^2+3u/4u^2+11-3=0

We add all the numbers together, and all the variables

(-32u)/4u^2+3u/4u^2+8=0

We multiply all the terms by the denominator


(-32u)+3u+8*4u^2=0

We add all the numbers together, and all the variables

3u+(-32u)+8*4u^2=0

Wy multiply elements

32u^2+3u+(-32u)=0

We get rid of parentheses

32u^2+3u-32u=0

We add all the numbers together, and all the variables

32u^2-29u=0

a = 32; b = -29; c = 0;
Δ = b2-4ac
Δ = -292-4·32·0
Δ = 841
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{841}=29$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-29)-29}{2*32}=\frac{0}{64}
=0
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-29)+29}{2*32}=\frac{58}{64}
=29/32
$