(4q+6)(5q-3)=0

Solution for (4q+6)(5q-3)=0 equation:

(4q+6)(5q-3)=0

We multiply parentheses ..

(+20q^2-12q+30q-18)=0

We get rid of parentheses

20q^2-12q+30q-18=0

We add all the numbers together, and all the variables

20q^2+18q-18=0

a = 20; b = 18; c = -18;
Δ = b2-4ac
Δ = 182-4·20·(-18)
Δ = 1764
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1764}=42$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-42}{2*20}=\frac{-60}{40}
=-1+1/2
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+42}{2*20}=\frac{24}{40}
=3/5
$