(m-3)(m+4)=8

Solution for (m-3)(m+4)=8 equation:

(m-3)(m+4)=8

We move all terms to the left:

(m-3)(m+4)-(8)=0

We multiply parentheses ..

(+m^2+4m-3m-12)-8=0

We get rid of parentheses

m^2+4m-3m-12-8=0

We add all the numbers together, and all the variables

m^2+m-20=0

a = 1; b = 1; c = -20;
Δ = b2-4ac
Δ = 12-4·1·(-20)
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-9}{2*1}=\frac{-10}{2}
=-5
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+9}{2*1}=\frac{8}{2}
=4
$