(q-7)(q+2)=0

Solution for (q-7)(q+2)=0 equation:

(q-7)(q+2)=0

We multiply parentheses ..

(+q^2+2q-7q-14)=0

We get rid of parentheses

q^2+2q-7q-14=0

We add all the numbers together, and all the variables

q^2-5q-14=0

a = 1; b = -5; c = -14;
Δ = b2-4ac
Δ = -52-4·1·(-14)
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-9}{2*1}=\frac{-4}{2}
=-2
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+9}{2*1}=\frac{14}{2}
=7
$