18+1/50q=-1/10q+210

Solution for 18+1/50q=-1/10q+210 equation:

18+1/50q=-1/10q+210

We move all terms to the left:

18+1/50q-(-1/10q+210)=0


Domain of the equation: 50q!=0

q!=0/50

q!=0

q∈R



Domain of the equation: 10q+210)!=0

q∈R


We get rid of parentheses

1/50q+1/10q-210+18=0

We calculate fractions


10q/500q^2+50q/500q^2-210+18=0

We add all the numbers together, and all the variables

10q/500q^2+50q/500q^2-192=0

We multiply all the terms by the denominator


10q+50q-192*500q^2=0

We add all the numbers together, and all the variables

60q-192*500q^2=0

Wy multiply elements

-96000q^2+60q=0

a = -96000; b = 60; c = 0;
Δ = b2-4ac
Δ = 602-4·(-96000)·0
Δ = 3600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3600}=60$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(60)-60}{2*-96000}=\frac{-120}{-192000}
=1/1600
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(60)+60}{2*-96000}=\frac{0}{-192000}
=0
$