-x^2+20x-1=0

Solution for -x^2+20x-1=0 equation:

-x^2+20x-1=0

We add all the numbers together, and all the variables

-1x^2+20x-1=0

a = -1; b = 20; c = -1;
Δ = b2-4ac
Δ = 202-4·(-1)·(-1)
Δ = 396
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{396}=\sqrt{36*11}=\sqrt{36}*\sqrt{11}=6\sqrt{11}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-6\sqrt{11}}{2*-1}=\frac{-20-6\sqrt{11}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+6\sqrt{11}}{2*-1}=\frac{-20+6\sqrt{11}}{-2}
$