8m+8(4+m)=(4+m)m

Solution for 8m+8(4+m)=(4+m)m equation:

8m+8(4+m)=(4+m)m

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses

8m+8m-((m+4)m)+32=0


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

(m+4)m

We multiply parentheses

m^2+4m

Back to the equation:

-(m^2+4m)


We add all the numbers together, and all the variables

16m-(m^2+4m)+32=0

We get rid of parentheses

-m^2+16m-4m+32=0

We add all the numbers together, and all the variables

-1m^2+12m+32=0

a = -1; b = 12; c = +32;
Δ = b2-4ac
Δ = 122-4·(-1)·32
Δ = 272
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:

$\sqrt{\Delta}=\sqrt{272}=\sqrt{16*17}=\sqrt{16}*\sqrt{17}=4\sqrt{17}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-4\sqrt{17}}{2*-1}=\frac{-12-4\sqrt{17}}{-2}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4\sqrt{17}}{2*-1}=\frac{-12+4\sqrt{17}}{-2}
$