2m^2-4m-80=0

Solution for 2m^2-4m-80=0 equation:

2m^2-4m-80=0

a = 2; b = -4; c = -80;
Δ = b2-4ac
Δ = -42-4·2·(-80)
Δ = 656
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:

$\sqrt{\Delta}=\sqrt{656}=\sqrt{16*41}=\sqrt{16}*\sqrt{41}=4\sqrt{41}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4\sqrt{41}}{2*2}=\frac{4-4\sqrt{41}}{4}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4\sqrt{41}}{2*2}=\frac{4+4\sqrt{41}}{4}
$