(q+5)(q-1)=27

Solution for (q+5)(q-1)=27 equation:

(q+5)(q-1)=27

We move all terms to the left:

(q+5)(q-1)-(27)=0

We multiply parentheses ..

(+q^2-1q+5q-5)-27=0

We get rid of parentheses

q^2-1q+5q-5-27=0

We add all the numbers together, and all the variables

q^2+4q-32=0

a = 1; b = 4; c = -32;
Δ = b2-4ac
Δ = 42-4·1·(-32)
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-12}{2*1}=\frac{-16}{2}
=-8
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+12}{2*1}=\frac{8}{2}
=4
$