4m-5=-9+3m(-5+2m)

Solution for 4m-5=-9+3m(-5+2m) equation:

4m-5=-9+3m(-5+2m)

We move all terms to the left:

4m-5-(-9+3m(-5+2m))=0

We add all the numbers together, and all the variables

4m-(-9+3m(2m-5))-5=0


We calculate terms in parentheses: -(-9+3m(2m-5)), so:

-9+3m(2m-5)

determiningTheFunctionDomain
3m(2m-5)-9

We multiply parentheses

6m^2-15m-9

Back to the equation:

-(6m^2-15m-9)


We get rid of parentheses

-6m^2+4m+15m+9-5=0

We add all the numbers together, and all the variables

-6m^2+19m+4=0

a = -6; b = 19; c = +4;
Δ = b2-4ac
Δ = 192-4·(-6)·4
Δ = 457
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{-(19)-\sqrt{457}}{2*-6}=\frac{-19-\sqrt{457}}{-12}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+\sqrt{457}}{2*-6}=\frac{-19+\sqrt{457}}{-12}
$