(4-m)(8+2/3m)=0

Solution for (4-m)(8+2/3m)=0 equation:

(4-m)(8+2/3m)=0


Domain of the equation: 3m)!=0

m!=0/1

m!=0

m∈R


We add all the numbers together, and all the variables

(-1m+4)(2/3m+8)=0

We multiply parentheses ..

(-2m^2-8m+8m+32)=0

We get rid of parentheses

-2m^2-8m+8m+32=0

We add all the numbers together, and all the variables

-2m^2+32=0

a = -2; b = 0; c = +32;
Δ = b2-4ac
Δ = 02-4·(-2)·32
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16}{2*-2}=\frac{-16}{-4}
=+4
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16}{2*-2}=\frac{16}{-4}
=-4
$