k^2-26k+3=0

Solution for k^2-26k+3=0 equation:

k^2-26k+3=0

a = 1; b = -26; c = +3;
Δ = b2-4ac
Δ = -262-4·1·3
Δ = 664
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{664}=\sqrt{4*166}=\sqrt{4}*\sqrt{166}=2\sqrt{166}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-26)-2\sqrt{166}}{2*1}=\frac{26-2\sqrt{166}}{2}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-26)+2\sqrt{166}}{2*1}=\frac{26+2\sqrt{166}}{2}
$