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

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

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

We move all terms to the left:

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


Domain of the equation: u!=0

u∈R


We add all the numbers together, and all the variables

u-4/u-3-((-11)+1)=0

We add all the numbers together, and all the variables

u-4/u+7=0

We multiply all the terms by the denominator


u*u+7*u-4=0

We add all the numbers together, and all the variables

7u+u*u-4=0

Wy multiply elements

u^2+7u-4=0

a = 1; b = 7; c = -4;
Δ = b2-4ac
Δ = 72-4·1·(-4)
Δ = 65
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-\sqrt{65}}{2*1}=\frac{-7-\sqrt{65}}{2}
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+\sqrt{65}}{2*1}=\frac{-7+\sqrt{65}}{2}
$