24d^2+64d+40=0

Solution for 24d^2+64d+40=0 equation:

24d^2+64d+40=0

a = 24; b = 64; c = +40;
Δ = b2-4ac
Δ = 642-4·24·40
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(64)-16}{2*24}=\frac{-80}{48}
=-1+2/3
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(64)+16}{2*24}=\frac{-48}{48}
=-1
$