(m+2)(m-3)=4

Solution for (m+2)(m-3)=4 equation:

(m+2)(m-3)=4

We move all terms to the left:

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

We multiply parentheses ..

(+m^2-3m+2m-6)-4=0

We get rid of parentheses

m^2-3m+2m-6-4=0

We add all the numbers together, and all the variables

m^2-1m-10=0

a = 1; b = -1; c = -10;
Δ = b2-4ac
Δ = -12-4·1·(-10)
Δ = 41
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-\sqrt{41}}{2*1}=\frac{1-\sqrt{41}}{2}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{41}}{2*1}=\frac{1+\sqrt{41}}{2}
$