(4q-9)(q-6)=0

Solution for (4q-9)(q-6)=0 equation:

(4q-9)(q-6)=0

We multiply parentheses ..

(+4q^2-24q-9q+54)=0

We get rid of parentheses

4q^2-24q-9q+54=0

We add all the numbers together, and all the variables

4q^2-33q+54=0

a = 4; b = -33; c = +54;
Δ = b2-4ac
Δ = -332-4·4·54
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-33)-15}{2*4}=\frac{18}{8}
=2+1/4
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-33)+15}{2*4}=\frac{48}{8}
=6
$