(7q+5)(q+4)=0

Solution for (7q+5)(q+4)=0 equation:

(7q+5)(q+4)=0

We multiply parentheses ..

(+7q^2+28q+5q+20)=0

We get rid of parentheses

7q^2+28q+5q+20=0

We add all the numbers together, and all the variables

7q^2+33q+20=0

a = 7; b = 33; c = +20;
Δ = b2-4ac
Δ = 332-4·7·20
Δ = 529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{529}=23$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(33)-23}{2*7}=\frac{-56}{14}
=-4
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(33)+23}{2*7}=\frac{-10}{14}
=-5/7
$