(9q+18)(2q+1)/q=3

Solution for (9q+18)(2q+1)/q=3 equation:

(9q+18)(2q+1)/q=3

We move all terms to the left:

(9q+18)(2q+1)/q-(3)=0


Domain of the equation: q!=0

q∈R


We multiply parentheses ..

(+18q^2+9q+36q+18)/q-3=0

We multiply all the terms by the denominator


(+18q^2+9q+36q+18)-3*q=0

We add all the numbers together, and all the variables

(+18q^2+9q+36q+18)-3q=0

We get rid of parentheses

18q^2+9q+36q-3q+18=0

We add all the numbers together, and all the variables

18q^2+42q+18=0

a = 18; b = 42; c = +18;
Δ = b2-4ac
Δ = 422-4·18·18
Δ = 468
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{468}=\sqrt{36*13}=\sqrt{36}*\sqrt{13}=6\sqrt{13}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-6\sqrt{13}}{2*18}=\frac{-42-6\sqrt{13}}{36}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+6\sqrt{13}}{2*18}=\frac{-42+6\sqrt{13}}{36}
$