m^2+20m-17=0

Solution for m^2+20m-17=0 equation:

m^2+20m-17=0

a = 1; b = 20; c = -17;
Δ = b2-4ac
Δ = 202-4·1·(-17)
Δ = 468
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:

$\sqrt{\Delta}=\sqrt{468}=\sqrt{36*13}=\sqrt{36}*\sqrt{13}=6\sqrt{13}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-6\sqrt{13}}{2*1}=\frac{-20-6\sqrt{13}}{2}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+6\sqrt{13}}{2*1}=\frac{-20+6\sqrt{13}}{2}
$