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

Solution for 16+m^2+16m-4=0 equation:

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

We add all the numbers together, and all the variables

m^2+16m+12=0

a = 1; b = 16; c = +12;
Δ = b2-4ac
Δ = 162-4·1·12
Δ = 208
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:

$\sqrt{\Delta}=\sqrt{208}=\sqrt{16*13}=\sqrt{16}*\sqrt{13}=4\sqrt{13}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-4\sqrt{13}}{2*1}=\frac{-16-4\sqrt{13}}{2}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+4\sqrt{13}}{2*1}=\frac{-16+4\sqrt{13}}{2}
$