(4q-1)(3q-8)=0

Solution for (4q-1)(3q-8)=0 equation:

(4q-1)(3q-8)=0

We multiply parentheses ..

(+12q^2-32q-3q+8)=0

We get rid of parentheses

12q^2-32q-3q+8=0

We add all the numbers together, and all the variables

12q^2-35q+8=0

a = 12; b = -35; c = +8;
Δ = b2-4ac
Δ = -352-4·12·8
Δ = 841
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{841}=29$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-35)-29}{2*12}=\frac{6}{24}
=1/4
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-35)+29}{2*12}=\frac{64}{24}
=2+2/3
$