-m^2-12m+4=0

Solution for -m^2-12m+4=0 equation:

-m^2-12m+4=0

We add all the numbers together, and all the variables

-1m^2-12m+4=0

a = -1; b = -12; c = +4;
Δ = b2-4ac
Δ = -122-4·(-1)·4
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{-(-12)-4\sqrt{10}}{2*-1}=\frac{12-4\sqrt{10}}{-2}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4\sqrt{10}}{2*-1}=\frac{12+4\sqrt{10}}{-2}
$