m^2+8m-1=0

Solution for m^2+8m-1=0 equation:

m^2+8m-1=0

a = 1; b = 8; c = -1;
Δ = b2-4ac
Δ = 82-4·1·(-1)
Δ = 68
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:

$\sqrt{\Delta}=\sqrt{68}=\sqrt{4*17}=\sqrt{4}*\sqrt{17}=2\sqrt{17}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-2\sqrt{17}}{2*1}=\frac{-8-2\sqrt{17}}{2}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+2\sqrt{17}}{2*1}=\frac{-8+2\sqrt{17}}{2}
$