(2q-3)q-4)=18

Solution for (2q-3)q-4)=18 equation:

(2q-3)q-4)=18

We move all terms to the left:

(2q-3)q-4)-(18)=0

We add all the numbers together, and all the variables

(2q-3)q=0

We multiply parentheses

2q^2-3q=0

a = 2; b = -3; c = 0;
Δ = b2-4ac
Δ = -32-4·2·0
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-3}{2*2}=\frac{0}{4}
=0
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+3}{2*2}=\frac{6}{4}
=1+1/2
$