2d^2+16=-33d

Solution for 2d^2+16=-33d equation:

2d^2+16=-33d

We move all terms to the left:

2d^2+16-(-33d)=0

We get rid of parentheses

2d^2+33d+16=0

a = 2; b = 33; c = +16;
Δ = b2-4ac
Δ = 332-4·2·16
Δ = 961
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{961}=31$
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(33)-31}{2*2}=\frac{-64}{4}
=-16
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(33)+31}{2*2}=\frac{-2}{4}
=-1/2
$