-9600+r^2+40r=0

Solution for -9600+r^2+40r=0 equation:

-9600+r^2+40r=0

a = 1; b = 40; c = -9600;
Δ = b2-4ac
Δ = 402-4·1·(-9600)
Δ = 40000
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{40000}=200$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-200}{2*1}=\frac{-240}{2}
=-120
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+200}{2*1}=\frac{160}{2}
=80
$