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

Solution for (m+3)-(m-6)m+2=m+2 equation:

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

We move all terms to the left:

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

We multiply parentheses

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

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

a = -1; b = 6; c = +3;
Δ = b2-4ac
Δ = 62-4·(-1)·3
Δ = 48
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:

$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-4\sqrt{3}}{2*-1}=\frac{-6-4\sqrt{3}}{-2}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+4\sqrt{3}}{2*-1}=\frac{-6+4\sqrt{3}}{-2}
$