(2q-3)(q-3)=18

Solution for (2q-3)(q-3)=18 equation:

(2q-3)(q-3)=18

We move all terms to the left:

(2q-3)(q-3)-(18)=0

We multiply parentheses ..

(+2q^2-6q-3q+9)-18=0

We get rid of parentheses

2q^2-6q-3q+9-18=0

We add all the numbers together, and all the variables

2q^2-9q-9=0

a = 2; b = -9; c = -9;
Δ = b2-4ac
Δ = -92-4·2·(-9)
Δ = 153
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:

$\sqrt{\Delta}=\sqrt{153}=\sqrt{9*17}=\sqrt{9}*\sqrt{17}=3\sqrt{17}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-3\sqrt{17}}{2*2}=\frac{9-3\sqrt{17}}{4}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+3\sqrt{17}}{2*2}=\frac{9+3\sqrt{17}}{4}
$