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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

m^2-1m-9=0

a = 1; b = -1; c = -9;
Δ = b2-4ac
Δ = -12-4·1·(-9)
Δ = 37
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{37}}{2*1}=\frac{1-\sqrt{37}}{2}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{37}}{2*1}=\frac{1+\sqrt{37}}{2}
$