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

Solution for (m-4)(m+3)=0 equation:

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

m^2-1m-12=0

a = 1; b = -1; c = -12;
Δ = b2-4ac
Δ = -12-4·1·(-12)
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-7}{2*1}=\frac{-6}{2}
=-3
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+7}{2*1}=\frac{8}{2}
=4
$