m^2+12m-16=0

Solution for m^2+12m-16=0 equation:

m^2+12m-16=0

a = 1; b = 12; c = -16;
Δ = b2-4ac
Δ = 122-4·1·(-16)
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{-(12)-4\sqrt{13}}{2*1}=\frac{-12-4\sqrt{13}}{2}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4\sqrt{13}}{2*1}=\frac{-12+4\sqrt{13}}{2}
$