1+20x-2x^2=0

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

1+20x-2x^2=0

a = -2; b = 20; c = +1;
Δ = b2-4ac
Δ = 202-4·(-2)·1
Δ = 408
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:

$\sqrt{\Delta}=\sqrt{408}=\sqrt{4*102}=\sqrt{4}*\sqrt{102}=2\sqrt{102}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-2\sqrt{102}}{2*-2}=\frac{-20-2\sqrt{102}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+2\sqrt{102}}{2*-2}=\frac{-20+2\sqrt{102}}{-4}
$