k^2-8k+3=0

Solution for k^2-8k+3=0 equation:

k^2-8k+3=0

a = 1; b = -8; c = +3;
Δ = b2-4ac
Δ = -82-4·1·3
Δ = 52
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:

$\sqrt{\Delta}=\sqrt{52}=\sqrt{4*13}=\sqrt{4}*\sqrt{13}=2\sqrt{13}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-2\sqrt{13}}{2*1}=\frac{8-2\sqrt{13}}{2}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+2\sqrt{13}}{2*1}=\frac{8+2\sqrt{13}}{2}
$