(8q-3)(q-2)=0

Solution for (8q-3)(q-2)=0 equation:

(8q-3)(q-2)=0

We multiply parentheses ..

(+8q^2-16q-3q+6)=0

We get rid of parentheses

8q^2-16q-3q+6=0

We add all the numbers together, and all the variables

8q^2-19q+6=0

a = 8; b = -19; c = +6;
Δ = b2-4ac
Δ = -192-4·8·6
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-13}{2*8}=\frac{6}{16}
=3/8
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+13}{2*8}=\frac{32}{16}
=2
$