m^2+16m-8=0

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

m^2+16m-8=0

a = 1; b = 16; c = -8;
Δ = b2-4ac
Δ = 162-4·1·(-8)
Δ = 288
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:

$\sqrt{\Delta}=\sqrt{288}=\sqrt{144*2}=\sqrt{144}*\sqrt{2}=12\sqrt{2}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-12\sqrt{2}}{2*1}=\frac{-16-12\sqrt{2}}{2}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+12\sqrt{2}}{2*1}=\frac{-16+12\sqrt{2}}{2}
$