-x^2+20x=2

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

-x^2+20x=2

We move all terms to the left:

-x^2+20x-(2)=0

We add all the numbers together, and all the variables

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

a = -1; b = 20; c = -2;
Δ = b2-4ac
Δ = 202-4·(-1)·(-2)
Δ = 392
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:

$\sqrt{\Delta}=\sqrt{392}=\sqrt{196*2}=\sqrt{196}*\sqrt{2}=14\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-14\sqrt{2}}{2*-1}=\frac{-20-14\sqrt{2}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+14\sqrt{2}}{2*-1}=\frac{-20+14\sqrt{2}}{-2}
$