m^2+8m-4=0

Solution for m^2+8m-4=0 equation:

m^2+8m-4=0

a = 1; b = 8; c = -4;
Δ = b2-4ac
Δ = 82-4·1·(-4)
Δ = 80
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:

$\sqrt{\Delta}=\sqrt{80}=\sqrt{16*5}=\sqrt{16}*\sqrt{5}=4\sqrt{5}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-4\sqrt{5}}{2*1}=\frac{-8-4\sqrt{5}}{2}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+4\sqrt{5}}{2*1}=\frac{-8+4\sqrt{5}}{2}
$