(4q-3)(2q-1)=0

Solution for (4q-3)(2q-1)=0 equation:

(4q-3)(2q-1)=0

We multiply parentheses ..

(+8q^2-4q-6q+3)=0

We get rid of parentheses

8q^2-4q-6q+3=0

We add all the numbers together, and all the variables

8q^2-10q+3=0

a = 8; b = -10; c = +3;
Δ = b2-4ac
Δ = -102-4·8·3
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2}{2*8}=\frac{8}{16}
=1/2
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2}{2*8}=\frac{12}{16}
=3/4
$