9d^2+20d+4=0

Solution for 9d^2+20d+4=0 equation:

9d^2+20d+4=0

a = 9; b = 20; c = +4;
Δ = b2-4ac
Δ = 202-4·9·4
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{-(20)-16}{2*9}=\frac{-36}{18}
=-2
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+16}{2*9}=\frac{-4}{18}
=-2/9
$