2m^2+16m-8=0

Solution for 2m^2+16m-8=0 equation:

2m^2+16m-8=0

a = 2; b = 16; c = -8;
Δ = b2-4ac
Δ = 162-4·2·(-8)
Δ = 320
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:

$\sqrt{\Delta}=\sqrt{320}=\sqrt{64*5}=\sqrt{64}*\sqrt{5}=8\sqrt{5}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-8\sqrt{5}}{2*2}=\frac{-16-8\sqrt{5}}{4}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+8\sqrt{5}}{2*2}=\frac{-16+8\sqrt{5}}{4}
$