11/b+2=3/5b+10

Solution for 11/b+2=3/5b+10 equation:

11/b+2=3/5b+10

We move all terms to the left:

11/b+2-(3/5b+10)=0


Domain of the equation: b!=0

b∈R



Domain of the equation: 5b+10)!=0

b∈R


We get rid of parentheses

11/b-3/5b-10+2=0

We calculate fractions


55b/5b^2+(-3b)/5b^2-10+2=0

We add all the numbers together, and all the variables

55b/5b^2+(-3b)/5b^2-8=0

We multiply all the terms by the denominator


55b+(-3b)-8*5b^2=0

Wy multiply elements

-40b^2+55b+(-3b)=0

We get rid of parentheses

-40b^2+55b-3b=0

We add all the numbers together, and all the variables

-40b^2+52b=0

a = -40; b = 52; c = 0;
Δ = b2-4ac
Δ = 522-4·(-40)·0
Δ = 2704
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2704}=52$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(52)-52}{2*-40}=\frac{-104}{-80}
=1+3/10
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(52)+52}{2*-40}=\frac{0}{-80}
=0
$