-16.1x^2+20x+50=0

Solution for -16.1x^2+20x+50=0 equation:

-16.1x^2+20x+50=0

a = -16.1; b = 20; c = +50;
Δ = b2-4ac
Δ = 202-4·(-16.1)·50
Δ = 3620
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:

$\sqrt{\Delta}=\sqrt{3620}=\sqrt{4*905}=\sqrt{4}*\sqrt{905}=2\sqrt{905}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-2\sqrt{905}}{2*-16.1}=\frac{-20-2\sqrt{905}}{-32.2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+2\sqrt{905}}{2*-16.1}=\frac{-20+2\sqrt{905}}{-32.2}
$