(q+5)(q-2)=0

Solution for (q+5)(q-2)=0 equation:

(q+5)(q-2)=0

We multiply parentheses ..

(+q^2-2q+5q-10)=0

We get rid of parentheses

q^2-2q+5q-10=0

We add all the numbers together, and all the variables

q^2+3q-10=0

a = 1; b = 3; c = -10;
Δ = b2-4ac
Δ = 32-4·1·(-10)
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{-(3)-7}{2*1}=\frac{-10}{2}
=-5
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+7}{2*1}=\frac{4}{2}
=2
$