(4q+3)(q+2)=0.

Solution for (4q+3)(q+2)=0. equation:

(4q+3)(q+2)=0.

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

4q^2+11q+6=0

a = 4; b = 11; c = +6;
Δ = b2-4ac
Δ = 112-4·4·6
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-5}{2*4}=\frac{-16}{8}
=-2
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+5}{2*4}=\frac{-6}{8}
=-3/4
$