10/3r+4/5r=403/20

Solution for 10/3r+4/5r=403/20 equation:

10/3r+4/5r=403/20

We move all terms to the left:

10/3r+4/5r-(403/20)=0


Domain of the equation: 3r!=0

r!=0/3

r!=0

r∈R



Domain of the equation: 5r!=0

r!=0/5

r!=0

r∈R


We add all the numbers together, and all the variables

10/3r+4/5r-(+403/20)=0

We get rid of parentheses

10/3r+4/5r-403/20=0

We calculate fractions


(-30225r^2)/600r^2+2000r/600r^2+480r/600r^2=0

We multiply all the terms by the denominator


(-30225r^2)+2000r+480r=0

We add all the numbers together, and all the variables

(-30225r^2)+2480r=0

We get rid of parentheses

-30225r^2+2480r=0

a = -30225; b = 2480; c = 0;
Δ = b2-4ac
Δ = 24802-4·(-30225)·0
Δ = 6150400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{6150400}=2480$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2480)-2480}{2*-30225}=\frac{-4960}{-60450}
=16/195
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2480)+2480}{2*-30225}=\frac{0}{-60450}
=0
$