4m^2+16m-16=0

Solution for 4m^2+16m-16=0 equation:

4m^2+16m-16=0

a = 4; b = 16; c = -16;
Δ = b2-4ac
Δ = 162-4·4·(-16)
Δ = 512
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:

$\sqrt{\Delta}=\sqrt{512}=\sqrt{256*2}=\sqrt{256}*\sqrt{2}=16\sqrt{2}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-16\sqrt{2}}{2*4}=\frac{-16-16\sqrt{2}}{8}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+16\sqrt{2}}{2*4}=\frac{-16+16\sqrt{2}}{8}
$