(q-7)(q+8)=0

Solution for (q-7)(q+8)=0 equation:

(q-7)(q+8)=0

We multiply parentheses ..

(+q^2+8q-7q-56)=0

We get rid of parentheses

q^2+8q-7q-56=0

We add all the numbers together, and all the variables

q^2+q-56=0

a = 1; b = 1; c = -56;
Δ = b2-4ac
Δ = 12-4·1·(-56)
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-15}{2*1}=\frac{-16}{2}
=-8
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+15}{2*1}=\frac{14}{2}
=7
$