(d-10)+(d/3)d=21

Solution for (d-10)+(d/3)d=21 equation:

(d-10)+(d/3)d=21

We move all terms to the left:

(d-10)+(d/3)d-(21)=0


Domain of the equation: 3)d!=0

d!=0/1

d!=0

d∈R


We add all the numbers together, and all the variables

(d-10)+(+d/3)d-21=0

We multiply parentheses

d^2+(d-10)-21=0

We get rid of parentheses

d^2+d-10-21=0

We add all the numbers together, and all the variables

d^2+d-31=0

a = 1; b = 1; c = -31;
Δ = b2-4ac
Δ = 12-4·1·(-31)
Δ = 125
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:

$\sqrt{\Delta}=\sqrt{125}=\sqrt{25*5}=\sqrt{25}*\sqrt{5}=5\sqrt{5}$
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-5\sqrt{5}}{2*1}=\frac{-1-5\sqrt{5}}{2}
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+5\sqrt{5}}{2*1}=\frac{-1+5\sqrt{5}}{2}
$