7/3u-5/3=5/2u+5

Solution for 7/3u-5/3=5/2u+5 equation:

7/3u-5/3=5/2u+5

We move all terms to the left:

7/3u-5/3-(5/2u+5)=0


Domain of the equation: 3u!=0

u!=0/3

u!=0

u∈R



Domain of the equation: 2u+5)!=0

u∈R


We get rid of parentheses

7/3u-5/2u-5-5/3=0

We calculate fractions


14u/54u^2+(-135u)/54u^2+(-10u)/54u^2-5=0

We multiply all the terms by the denominator


14u+(-135u)+(-10u)-5*54u^2=0

Wy multiply elements

-270u^2+14u+(-135u)+(-10u)=0

We get rid of parentheses

-270u^2+14u-135u-10u=0

We add all the numbers together, and all the variables

-270u^2-131u=0

a = -270; b = -131; c = 0;
Δ = b2-4ac
Δ = -1312-4·(-270)·0
Δ = 17161
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{17161}=131$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-131)-131}{2*-270}=\frac{0}{-540}
=0
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-131)+131}{2*-270}=\frac{262}{-540}
=-131/270
$