3d^2+d=2

Solution for 3d^2+d=2 equation:

3d^2+d=2

We move all terms to the left:

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

a = 3; b = 1; c = -2;
Δ = b2-4ac
Δ = 12-4·3·(-2)
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-5}{2*3}=\frac{-6}{6}
=-1
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+5}{2*3}=\frac{4}{6}
=2/3
$