d+(1/3)d+100+1.25d-10=400

Solution for d+(1/3)d+100+1.25d-10=400 equation:

d+(1/3)d+100+1.25d-10=400

We move all terms to the left:

d+(1/3)d+100+1.25d-10-(400)=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+(+1/3)d+1.25d+100-10-400=0

We add all the numbers together, and all the variables

2.25d+(+1/3)d-310=0

We multiply parentheses

d^2+2.25d-310=0

a = 1; b = 2.25; c = -310;
Δ = b2-4ac
Δ = 2.252-4·1·(-310)
Δ = 1245.0625
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{-(2.25)-\sqrt{1245.0625}}{2*1}=\frac{-2.25-\sqrt{1245.0625}}{2}
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2.25)+\sqrt{1245.0625}}{2*1}=\frac{-2.25+\sqrt{1245.0625}}{2}
$