3d2+d=-2d

Solution for 3d2+d=-2d equation:

3d^2+d=-2d

We move all terms to the left:

3d^2+d-(-2d)=0

We get rid of parentheses

3d^2+d+2d=0

We add all the numbers together, and all the variables

3d^2+3d=0

a = 3; b = 3; c = 0;
Δ = b2-4ac
Δ = 32-4·3·0
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3}{2*3}=\frac{-6}{6}
=-1
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3}{2*3}=\frac{0}{6}
=0
$