18=9q(9-q)

Solution for 18=9q(9-q) equation:

18=9q(9-q)

We move all terms to the left:

18-(9q(9-q))=0

We add all the numbers together, and all the variables

-(9q(-1q+9))+18=0


We calculate terms in parentheses: -(9q(-1q+9)), so:

9q(-1q+9)

We multiply parentheses

-9q^2+81q

Back to the equation:

-(-9q^2+81q)


We get rid of parentheses

9q^2-81q+18=0

a = 9; b = -81; c = +18;
Δ = b2-4ac
Δ = -812-4·9·18
Δ = 5913
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:

$\sqrt{\Delta}=\sqrt{5913}=\sqrt{81*73}=\sqrt{81}*\sqrt{73}=9\sqrt{73}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-81)-9\sqrt{73}}{2*9}=\frac{81-9\sqrt{73}}{18}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-81)+9\sqrt{73}}{2*9}=\frac{81+9\sqrt{73}}{18}
$