3m+2m(m+2)=20-(2m+5)

Solution for 3m+2m(m+2)=20-(2m+5) equation:

3m+2m(m+2)=20-(2m+5)

We move all terms to the left:

3m+2m(m+2)-(20-(2m+5))=0

We multiply parentheses

2m^2+3m+4m-(20-(2m+5))=0


We calculate terms in parentheses: -(20-(2m+5)), so:

20-(2m+5)

determiningTheFunctionDomain
-(2m+5)+20

We get rid of parentheses

-2m-5+20

We add all the numbers together, and all the variables

-2m+15

Back to the equation:

-(-2m+15)


We add all the numbers together, and all the variables

2m^2+7m-(-2m+15)=0

We get rid of parentheses

2m^2+7m+2m-15=0

We add all the numbers together, and all the variables

2m^2+9m-15=0

a = 2; b = 9; c = -15;
Δ = b2-4ac
Δ = 92-4·2·(-15)
Δ = 201
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{-(9)-\sqrt{201}}{2*2}=\frac{-9-\sqrt{201}}{4}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+\sqrt{201}}{2*2}=\frac{-9+\sqrt{201}}{4}
$