-(7/4u+12)+1=4/u+3

Solution for -(7/4u+12)+1=4/u+3 equation:

-(7/4u+12)+1=4/u+3

We move all terms to the left:

-(7/4u+12)+1-(4/u+3)=0


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

u∈R



Domain of the equation: u+3)!=0

u∈R


We get rid of parentheses

-7/4u-4/u-12-3+1=0

We calculate fractions


(-7u)/4u^2+(-16u)/4u^2-12-3+1=0

We add all the numbers together, and all the variables

(-7u)/4u^2+(-16u)/4u^2-14=0

We multiply all the terms by the denominator


(-7u)+(-16u)-14*4u^2=0

Wy multiply elements

-56u^2+(-7u)+(-16u)=0

We get rid of parentheses

-56u^2-7u-16u=0

We add all the numbers together, and all the variables

-56u^2-23u=0

a = -56; b = -23; c = 0;
Δ = b2-4ac
Δ = -232-4·(-56)·0
Δ = 529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{529}=23$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-23)-23}{2*-56}=\frac{0}{-112}
=0
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-23)+23}{2*-56}=\frac{46}{-112}
=-23/56
$