1/3q+2/5=1/5q+10

Solution for 1/3q+2/5=1/5q+10 equation:

1/3q+2/5=1/5q+10

We move all terms to the left:

1/3q+2/5-(1/5q+10)=0


Domain of the equation: 3q!=0

q!=0/3

q!=0

q∈R



Domain of the equation: 5q+10)!=0

q∈R


We get rid of parentheses

1/3q-1/5q-10+2/5=0

We calculate fractions


125q/375q^2+(-3q)/375q^2+6q/375q^2-10=0

We multiply all the terms by the denominator


125q+(-3q)+6q-10*375q^2=0

We add all the numbers together, and all the variables

131q+(-3q)-10*375q^2=0

Wy multiply elements

-3750q^2+131q+(-3q)=0

We get rid of parentheses

-3750q^2+131q-3q=0

We add all the numbers together, and all the variables

-3750q^2+128q=0

a = -3750; b = 128; c = 0;
Δ = b2-4ac
Δ = 1282-4·(-3750)·0
Δ = 16384
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16384}=128$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(128)-128}{2*-3750}=\frac{-256}{-7500}
=64/1875
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(128)+128}{2*-3750}=\frac{0}{-7500}
=0
$