4(d+2)-2d=3d(2+d)

Solution for 4(d+2)-2d=3d(2+d) equation:

4(d+2)-2d=3d(2+d)

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We add all the numbers together, and all the variables

-2d+4(d+2)-(3d(d+2))=0

We multiply parentheses

-2d+4d-(3d(d+2))+8=0


We calculate terms in parentheses: -(3d(d+2)), so:

3d(d+2)

We multiply parentheses

3d^2+6d

Back to the equation:

-(3d^2+6d)


We add all the numbers together, and all the variables

2d-(3d^2+6d)+8=0

We get rid of parentheses

-3d^2+2d-6d+8=0

We add all the numbers together, and all the variables

-3d^2-4d+8=0

a = -3; b = -4; c = +8;
Δ = b2-4ac
Δ = -42-4·(-3)·8
Δ = 112
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:

$\sqrt{\Delta}=\sqrt{112}=\sqrt{16*7}=\sqrt{16}*\sqrt{7}=4\sqrt{7}$
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4\sqrt{7}}{2*-3}=\frac{4-4\sqrt{7}}{-6}
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4\sqrt{7}}{2*-3}=\frac{4+4\sqrt{7}}{-6}
$