10d2+33d+20=0

Solution for 10d2+33d+20=0 equation:

10d^2+33d+20=0

a = 10; b = 33; c = +20;
Δ = b2-4ac
Δ = 332-4·10·20
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(33)-17}{2*10}=\frac{-50}{20}
=-2+1/2
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(33)+17}{2*10}=\frac{-16}{20}
=-4/5
$