3m+108=m^2

Solution for 3m+108=m^2 equation:

3m+108=m^2

We move all terms to the left:

3m+108-(m^2)=0

determiningTheFunctionDomain
-m^2+3m+108=0

We add all the numbers together, and all the variables

-1m^2+3m+108=0

a = -1; b = 3; c = +108;
Δ = b2-4ac
Δ = 32-4·(-1)·108
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{441}=21$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-21}{2*-1}=\frac{-24}{-2}
=+12
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+21}{2*-1}=\frac{18}{-2}
=-9
$