k^2-32k+11=0

Solution for k^2-32k+11=0 equation:

k^2-32k+11=0

a = 1; b = -32; c = +11;
Δ = b2-4ac
Δ = -322-4·1·11
Δ = 980
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:

$\sqrt{\Delta}=\sqrt{980}=\sqrt{196*5}=\sqrt{196}*\sqrt{5}=14\sqrt{5}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-14\sqrt{5}}{2*1}=\frac{32-14\sqrt{5}}{2}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+14\sqrt{5}}{2*1}=\frac{32+14\sqrt{5}}{2}
$