b-5=3/4b

Solution for b-5=3/4b equation:

b-5=3/4b

We move all terms to the left:

b-5-(3/4b)=0


Domain of the equation: 4b)!=0

b!=0/1

b!=0

b∈R


We add all the numbers together, and all the variables

b-(+3/4b)-5=0

We get rid of parentheses

b-3/4b-5=0

We multiply all the terms by the denominator


b*4b-5*4b-3=0

Wy multiply elements

4b^2-20b-3=0

a = 4; b = -20; c = -3;
Δ = b2-4ac
Δ = -202-4·4·(-3)
Δ = 448
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:

$\sqrt{\Delta}=\sqrt{448}=\sqrt{64*7}=\sqrt{64}*\sqrt{7}=8\sqrt{7}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-8\sqrt{7}}{2*4}=\frac{20-8\sqrt{7}}{8}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+8\sqrt{7}}{2*4}=\frac{20+8\sqrt{7}}{8}
$