2(20-3m)=m^2

Solution for 2(20-3m)=m^2 equation:

2(20-3m)=m^2

We move all terms to the left:

2(20-3m)-(m^2)=0

determiningTheFunctionDomain
-m^2+2(20-3m)=0

We add all the numbers together, and all the variables

-m^2+2(-3m+20)=0

We add all the numbers together, and all the variables

-1m^2+2(-3m+20)=0

We multiply parentheses

-1m^2-6m+40=0

a = -1; b = -6; c = +40;
Δ = b2-4ac
Δ = -62-4·(-1)·40
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-14}{2*-1}=\frac{-8}{-2}
=+4
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+14}{2*-1}=\frac{20}{-2}
=-10
$