-5x^2+20x+1=0

Solution for -5x^2+20x+1=0 equation:

-5x^2+20x+1=0

a = -5; b = 20; c = +1;
Δ = b2-4ac
Δ = 202-4·(-5)·1
Δ = 420
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:

$\sqrt{\Delta}=\sqrt{420}=\sqrt{4*105}=\sqrt{4}*\sqrt{105}=2\sqrt{105}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-2\sqrt{105}}{2*-5}=\frac{-20-2\sqrt{105}}{-10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+2\sqrt{105}}{2*-5}=\frac{-20+2\sqrt{105}}{-10}
$