(3m+2)(m-3)-(2m-1)(m+3)=0

Solution for (3m+2)(m-3)-(2m-1)(m+3)=0 equation:

(3m+2)(m-3)-(2m-1)(m+3)=0

We multiply parentheses ..

(+3m^2-9m+2m-6)-(2m-1)(m+3)=0

We get rid of parentheses

3m^2-9m+2m-(2m-1)(m+3)-6=0

We multiply parentheses ..

3m^2-(+2m^2+6m-1m-3)-9m+2m-6=0

We add all the numbers together, and all the variables

3m^2-(+2m^2+6m-1m-3)-7m-6=0

We get rid of parentheses

3m^2-2m^2-6m+1m-7m+3-6=0

We add all the numbers together, and all the variables

m^2-12m-3=0

a = 1; b = -12; c = -3;
Δ = b2-4ac
Δ = -122-4·1·(-3)
Δ = 156
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{156}=\sqrt{4*39}=\sqrt{4}*\sqrt{39}=2\sqrt{39}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-2\sqrt{39}}{2*1}=\frac{12-2\sqrt{39}}{2}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+2\sqrt{39}}{2*1}=\frac{12+2\sqrt{39}}{2}
$