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)(-1m+2))=0

We multiply parentheses ..

4m^2-((-1m^2+2m-4m+8))=0


We calculate terms in parentheses: -((-1m^2+2m-4m+8)), so:

(-1m^2+2m-4m+8)

We get rid of parentheses

-1m^2+2m-4m+8

We add all the numbers together, and all the variables

-1m^2-2m+8

Back to the equation:

-(-1m^2-2m+8)


We get rid of parentheses

4m^2+1m^2+2m-8=0

We add all the numbers together, and all the variables

5m^2+2m-8=0

a = 5; b = 2; c = -8;
Δ = b2-4ac
Δ = 22-4·5·(-8)
Δ = 164
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:

$\sqrt{\Delta}=\sqrt{164}=\sqrt{4*41}=\sqrt{4}*\sqrt{41}=2\sqrt{41}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{41}}{2*5}=\frac{-2-2\sqrt{41}}{10}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{41}}{2*5}=\frac{-2+2\sqrt{41}}{10}
$