4/4b+5=2/b-5

Solution for 4/4b+5=2/b-5 equation:

4/4b+5=2/b-5

We move all terms to the left:

4/4b+5-(2/b-5)=0


Domain of the equation: 4b!=0

b!=0/4

b!=0

b∈R



Domain of the equation: b-5)!=0

b∈R


We get rid of parentheses

4/4b-2/b+5+5=0

We calculate fractions


4b/4b^2+(-8b)/4b^2+5+5=0

We add all the numbers together, and all the variables

4b/4b^2+(-8b)/4b^2+10=0

We multiply all the terms by the denominator


4b+(-8b)+10*4b^2=0

Wy multiply elements

40b^2+4b+(-8b)=0

We get rid of parentheses

40b^2+4b-8b=0

We add all the numbers together, and all the variables

40b^2-4b=0

a = 40; b = -4; c = 0;
Δ = b2-4ac
Δ = -42-4·40·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4}{2*40}=\frac{0}{80}
=0
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4}{2*40}=\frac{8}{80}
=1/10
$