q/5q+3=2q+5

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

q/5q+3=2q+5

We move all terms to the left:

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


Domain of the equation: 5q!=0

q!=0/5

q!=0

q∈R


We get rid of parentheses

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

We multiply all the terms by the denominator


q-2q*5q-5*5q+3*5q=0

Wy multiply elements

-10q^2+q-25q+15q=0

We add all the numbers together, and all the variables

-10q^2-9q=0

a = -10; b = -9; c = 0;
Δ = b2-4ac
Δ = -92-4·(-10)·0
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-9}{2*-10}=\frac{0}{-20}
=0
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+9}{2*-10}=\frac{18}{-20}
=-9/10
$