(24-m)(9+3/4m)=0

Solution for (24-m)(9+3/4m)=0 equation:

(24-m)(9+3/4m)=0


Domain of the equation: 4m)!=0

m!=0/1

m!=0

m∈R


We add all the numbers together, and all the variables

(-1m+24)(3/4m+9)=0

We multiply parentheses ..

(-3m^2-9m+72m+216)=0

We get rid of parentheses

-3m^2-9m+72m+216=0

We add all the numbers together, and all the variables

-3m^2+63m+216=0

a = -3; b = 63; c = +216;
Δ = b2-4ac
Δ = 632-4·(-3)·216
Δ = 6561
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{6561}=81$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(63)-81}{2*-3}=\frac{-144}{-6}
=+24
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(63)+81}{2*-3}=\frac{18}{-6}
=-3
$