(w+3)(w-8)=0

Solution for (w+3)(w-8)=0 equation:

(w+3)(w-8)=0

We multiply parentheses ..

(+w^2-8w+3w-24)=0

We get rid of parentheses

w^2-8w+3w-24=0

We add all the numbers together, and all the variables

w^2-5w-24=0

a = 1; b = -5; c = -24;
Δ = b2-4ac
Δ = -52-4·1·(-24)
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-11}{2*1}=\frac{-6}{2}
=-3
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+11}{2*1}=\frac{16}{2}
=8
$