4m^2=(m+4)(2+m)

Solution for 4m^2=(m+4)(2+m) equation:

4m^2=(m+4)(2+m)

We move all terms to the left:

4m^2-((m+4)(2+m))=0

We add all the numbers together, and all the variables

4m^2-((m+4)(m+2))=0

We multiply parentheses ..

4m^2-((+m^2+2m+4m+8))=0


We calculate terms in parentheses: -((+m^2+2m+4m+8)), so:

(+m^2+2m+4m+8)

We get rid of parentheses

m^2+2m+4m+8

We add all the numbers together, and all the variables

m^2+6m+8

Back to the equation:

-(m^2+6m+8)


We get rid of parentheses

4m^2-m^2-6m-8=0

We add all the numbers together, and all the variables

3m^2-6m-8=0

a = 3; b = -6; c = -8;
Δ = b2-4ac
Δ = -62-4·3·(-8)
Δ = 132
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{132}=\sqrt{4*33}=\sqrt{4}*\sqrt{33}=2\sqrt{33}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{33}}{2*3}=\frac{6-2\sqrt{33}}{6}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{33}}{2*3}=\frac{6+2\sqrt{33}}{6}
$