-10x^2+20x+80=0

Solution for -10x^2+20x+80=0 equation:

-10x^2+20x+80=0

a = -10; b = 20; c = +80;
Δ = b2-4ac
Δ = 202-4·(-10)·80
Δ = 3600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3600}=60$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-60}{2*-10}=\frac{-80}{-20}
=+4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+60}{2*-10}=\frac{40}{-20}
=-2
$