(D^2+3D+4)y=0

Solution for (D^2+3D+4)y=0 equation:

(^2+3+4)D=0

We multiply parentheses

D^2+3D+4D=0

We add all the numbers together, and all the variables

D^2+7D=0

a = 1; b = 7; c = 0;
Δ = b2-4ac
Δ = 72-4·1·0
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-7}{2*1}=\frac{-14}{2}
=-7
$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+7}{2*1}=\frac{0}{2}
=0
$