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

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

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

We multiply parentheses ..

(+q^2+4q-3q-12)=0

We get rid of parentheses

q^2+4q-3q-12=0

We add all the numbers together, and all the variables

q^2+q-12=0

a = 1; b = 1; c = -12;
Δ = b2-4ac
Δ = 12-4·1·(-12)
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-7}{2*1}=\frac{-8}{2}
=-4
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+7}{2*1}=\frac{6}{2}
=3
$