b=1400/3b

Solution for b=1400/3b equation:

b=1400/3b

We move all terms to the left:

b-(1400/3b)=0


Domain of the equation: 3b)!=0

b!=0/1

b!=0

b∈R


We add all the numbers together, and all the variables

b-(+1400/3b)=0

We get rid of parentheses

b-1400/3b=0

We multiply all the terms by the denominator


b*3b-1400=0

Wy multiply elements

3b^2-1400=0

a = 3; b = 0; c = -1400;
Δ = b2-4ac
Δ = 02-4·3·(-1400)
Δ = 16800
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{16800}=\sqrt{400*42}=\sqrt{400}*\sqrt{42}=20\sqrt{42}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-20\sqrt{42}}{2*3}=\frac{0-20\sqrt{42}}{6}
=-\frac{20\sqrt{42}}{6}
=-\frac{10\sqrt{42}}{3}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+20\sqrt{42}}{2*3}=\frac{0+20\sqrt{42}}{6}
=\frac{20\sqrt{42}}{6}
=\frac{10\sqrt{42}}{3}
$