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

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

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

m^2-5m-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:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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