D^3+7d^2+19d+13=0

Solution for D^3+7d^2+19d+13=0 equation:

^3+7D^2+19D+13=0

We add all the numbers together, and all the variables

7D^2+19D=0

a = 7; b = 19; c = 0;
Δ = b2-4ac
Δ = 192-4·7·0
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-19}{2*7}=\frac{-38}{14}
=-2+5/7
$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+19}{2*7}=\frac{0}{14}
=0
$