D(d-1)+4d=d(d-8)+5

Solution for D(d-1)+4d=d(d-8)+5 equation:

(D-1)+4D=D(D-8)+5

We move all terms to the left:

(D-1)+4D-(D(D-8)+5)=0

We add all the numbers together, and all the variables

4D+(D-1)-(D(D-8)+5)=0

We get rid of parentheses

4D+D-(D(D-8)+5)-1=0


We calculate terms in parentheses: -(D(D-8)+5), so:

D(D-8)+5

We multiply parentheses

D^2-8D+5

Back to the equation:

-(D^2-8D+5)


We add all the numbers together, and all the variables

5D-(D^2-8D+5)-1=0

We get rid of parentheses

-D^2+5D+8D-5-1=0

We add all the numbers together, and all the variables

-1D^2+13D-6=0

a = -1; b = 13; c = -6;
Δ = b2-4ac
Δ = 132-4·(-1)·(-6)
Δ = 145
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-\sqrt{145}}{2*-1}=\frac{-13-\sqrt{145}}{-2}
$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+\sqrt{145}}{2*-1}=\frac{-13+\sqrt{145}}{-2}
$