K-x^2+6x+1=0

Solution for K-x^2+6x+1=0 equation:

-K^2+6K+1=0

We add all the numbers together, and all the variables

-1K^2+6K+1=0

a = -1; b = 6; c = +1;
Δ = b2-4ac
Δ = 62-4·(-1)·1
Δ = 40
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:

$\sqrt{\Delta}=\sqrt{40}=\sqrt{4*10}=\sqrt{4}*\sqrt{10}=2\sqrt{10}$
$K_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{10}}{2*-1}=\frac{-6-2\sqrt{10}}{-2}
$
$K_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{10}}{2*-1}=\frac{-6+2\sqrt{10}}{-2}
$