6/5r+1=4/3r=-373/45

Solution for 6/5r+1=4/3r=-373/45 equation:

6/5r+1=4/3r=-373/45

We move all terms to the left:

6/5r+1-(4/3r)=0


Domain of the equation: 5r!=0

r!=0/5

r!=0

r∈R



Domain of the equation: 3r)!=0

r!=0/1

r!=0

r∈R


We add all the numbers together, and all the variables

6/5r-(+4/3r)+1=0

We get rid of parentheses

6/5r-4/3r+1=0

We calculate fractions


18r/15r^2+(-20r)/15r^2+1=0

We multiply all the terms by the denominator


18r+(-20r)+1*15r^2=0

Wy multiply elements

15r^2+18r+(-20r)=0

We get rid of parentheses

15r^2+18r-20r=0

We add all the numbers together, and all the variables

15r^2-2r=0

a = 15; b = -2; c = 0;
Δ = b2-4ac
Δ = -22-4·15·0
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2}{2*15}=\frac{0}{30}
=0
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2}{2*15}=\frac{4}{30}
=2/15
$