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

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

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

We move all terms to the left:

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


Domain of the equation: 5b!=0

b!=0/5

b!=0

b∈R



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

b∈R


We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We multiply all the terms by the denominator


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

We add all the numbers together, and all the variables

-10b*5b+2*5b+8=0

Wy multiply elements

-50b^2+10b+8=0

a = -50; b = 10; c = +8;
Δ = b2-4ac
Δ = 102-4·(-50)·8
Δ = 1700
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:

$\sqrt{\Delta}=\sqrt{1700}=\sqrt{100*17}=\sqrt{100}*\sqrt{17}=10\sqrt{17}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-10\sqrt{17}}{2*-50}=\frac{-10-10\sqrt{17}}{-100}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+10\sqrt{17}}{2*-50}=\frac{-10+10\sqrt{17}}{-100}
$