m^2+16m=4

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

m^2+16m=4

We move all terms to the left:

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

a = 1; b = 16; c = -4;
Δ = b2-4ac
Δ = 162-4·1·(-4)
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{-(16)-4\sqrt{17}}{2*1}=\frac{-16-4\sqrt{17}}{2}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+4\sqrt{17}}{2*1}=\frac{-16+4\sqrt{17}}{2}
$