-5r^2+40r+400=0

Solution for -5r^2+40r+400=0 equation:

-5r^2+40r+400=0

a = -5; b = 40; c = +400;
Δ = b2-4ac
Δ = 402-4·(-5)·400
Δ = 9600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{9600}=\sqrt{1600*6}=\sqrt{1600}*\sqrt{6}=40\sqrt{6}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-40\sqrt{6}}{2*-5}=\frac{-40-40\sqrt{6}}{-10}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+40\sqrt{6}}{2*-5}=\frac{-40+40\sqrt{6}}{-10}
$