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

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

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

We move all terms to the left:

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

We multiply parentheses ..

2(+m^2+5m+5m+25)-20=0

We multiply parentheses

2m^2+10m+10m+50-20=0

We add all the numbers together, and all the variables

2m^2+20m+30=0

a = 2; b = 20; c = +30;
Δ = b2-4ac
Δ = 202-4·2·30
Δ = 160
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:

$\sqrt{\Delta}=\sqrt{160}=\sqrt{16*10}=\sqrt{16}*\sqrt{10}=4\sqrt{10}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{10}}{2*2}=\frac{-20-4\sqrt{10}}{4}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{10}}{2*2}=\frac{-20+4\sqrt{10}}{4}
$