(d/5)+3D-10=(1/3)d+33

Solution for (d/5)+3D-10=(1/3)d+33 equation:

(d/5)+3-10=(1/3)d+33

We move all terms to the left:

(d/5)+3-10-((1/3)d+33)=0


Domain of the equation: 3)d+33)!=0

d!=0/1

d!=0

d∈R


We add all the numbers together, and all the variables

(+d/5)-((+1/3)d+33)+3-10=0

We add all the numbers together, and all the variables

(+d/5)-((+1/3)d+33)-7=0

We get rid of parentheses

d/5-((+1/3)d+33)-7=0

We calculate fractions


3d^2/15d+()/15d-7=0

We multiply all the terms by the denominator


3d^2-7*15d+()=0

We add all the numbers together, and all the variables

3d^2-7*15d=0

Wy multiply elements

3d^2-105d=0

a = 3; b = -105; c = 0;
Δ = b2-4ac
Δ = -1052-4·3·0
Δ = 11025
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{11025}=105$
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-105)-105}{2*3}=\frac{0}{6}
=0
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-105)+105}{2*3}=\frac{210}{6}
=35
$