m^2+26m-20=0

Solution for m^2+26m-20=0 equation:

m^2+26m-20=0

a = 1; b = 26; c = -20;
Δ = b2-4ac
Δ = 262-4·1·(-20)
Δ = 756
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:

$\sqrt{\Delta}=\sqrt{756}=\sqrt{36*21}=\sqrt{36}*\sqrt{21}=6\sqrt{21}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(26)-6\sqrt{21}}{2*1}=\frac{-26-6\sqrt{21}}{2}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(26)+6\sqrt{21}}{2*1}=\frac{-26+6\sqrt{21}}{2}
$