(2D^2+3D+4)=x^2-2x

Solution for (2D^2+3D+4)=x^2-2x equation:

(2^2+3+4)=D^2-2D

We move all terms to the left:

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

We add all the numbers together, and all the variables

-(D^2-2D)+11=0

We get rid of parentheses

-D^2+2D+11=0

We add all the numbers together, and all the variables

-1D^2+2D+11=0

a = -1; b = 2; c = +11;
Δ = b2-4ac
Δ = 22-4·(-1)·11
Δ = 48
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$
$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-4\sqrt{3}}{2*-1}=\frac{-2-4\sqrt{3}}{-2}
$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+4\sqrt{3}}{2*-1}=\frac{-2+4\sqrt{3}}{-2}
$