m^-10m+24=(m-12)(m+2)

Solution for m^-10m+24=(m-12)(m+2) equation:

m^-10m+24=(m-12)(m+2)

We move all terms to the left:

m^-10m+24-((m-12)(m+2))=0

We add all the numbers together, and all the variables

-9m-((m-12)(m+2))+24=0

We multiply parentheses ..

-((+m^2+2m-12m-24))-9m+24=0


We calculate terms in parentheses: -((+m^2+2m-12m-24)), so:

(+m^2+2m-12m-24)

We get rid of parentheses

m^2+2m-12m-24

We add all the numbers together, and all the variables

m^2-10m-24

Back to the equation:

-(m^2-10m-24)


We add all the numbers together, and all the variables

-9m-(m^2-10m-24)+24=0

We get rid of parentheses

-m^2-9m+10m+24+24=0

We add all the numbers together, and all the variables

-1m^2+m+48=0

a = -1; b = 1; c = +48;
Δ = b2-4ac
Δ = 12-4·(-1)·48
Δ = 193
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{193}}{2*-1}=\frac{-1-\sqrt{193}}{-2}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{193}}{2*-1}=\frac{-1+\sqrt{193}}{-2}
$