(q-2)(q-6)=0

Solution for (q-2)(q-6)=0 equation:

(q-2)(q-6)=0

We multiply parentheses ..

(+q^2-6q-2q+12)=0

We get rid of parentheses

q^2-6q-2q+12=0

We add all the numbers together, and all the variables

q^2-8q+12=0

a = 1; b = -8; c = +12;
Δ = b2-4ac
Δ = -82-4·1·12
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-4}{2*1}=\frac{4}{2}
=2
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+4}{2*1}=\frac{12}{2}
=6
$