24r^2+40r=0

Solution for 24r^2+40r=0 equation:

24r^2+40r=0

a = 24; b = 40; c = 0;
Δ = b2-4ac
Δ = 402-4·24·0
Δ = 1600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1600}=40$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-40}{2*24}=\frac{-80}{48}
=-1+2/3
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+40}{2*24}=\frac{0}{48}
=0
$