d(d-1)=(2-d)+(d+4)

Solution for d(d-1)=(2-d)+(d+4) equation:

d(d-1)=(2-d)+(d+4)

We move all terms to the left:

d(d-1)-((2-d)+(d+4))=0

We add all the numbers together, and all the variables

d(d-1)-((-1d+2)+(d+4))=0

We multiply parentheses

d^2-1d-((-1d+2)+(d+4))=0


We calculate terms in parentheses: -((-1d+2)+(d+4)), so:

(-1d+2)+(d+4)

We get rid of parentheses

-1d+d+2+4

We add all the numbers together, and all the variables

6

Back to the equation:

-(6)


a = 1; b = -1; c = -6;
Δ = b2-4ac
Δ = -12-4·1·(-6)
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*1}=\frac{-4}{2}
=-2
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+5}{2*1}=\frac{6}{2}
=3
$