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

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

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

We move all terms to the left:

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

We multiply parentheses

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

We get rid of parentheses

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


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

4m(m+4)+4

We multiply parentheses

4m^2+16m+4

Back to the equation:

-(4m^2+16m+4)


We add all the numbers together, and all the variables

2m-(4m^2+16m+4)-1=0

We get rid of parentheses

-4m^2+2m-16m-4-1=0

We add all the numbers together, and all the variables

-4m^2-14m-5=0

a = -4; b = -14; c = -5;
Δ = b2-4ac
Δ = -142-4·(-4)·(-5)
Δ = 116
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:

$\sqrt{\Delta}=\sqrt{116}=\sqrt{4*29}=\sqrt{4}*\sqrt{29}=2\sqrt{29}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-2\sqrt{29}}{2*-4}=\frac{14-2\sqrt{29}}{-8}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+2\sqrt{29}}{2*-4}=\frac{14+2\sqrt{29}}{-8}
$