1/5b+3b=2b+42

Solution for 1/5b+3b=2b+42 equation:

1/5b+3b=2b+42

We move all terms to the left:

1/5b+3b-(2b+42)=0


Domain of the equation: 5b!=0

b!=0/5

b!=0

b∈R


We add all the numbers together, and all the variables

3b+1/5b-(2b+42)=0

We get rid of parentheses

3b+1/5b-2b-42=0

We multiply all the terms by the denominator


3b*5b-2b*5b-42*5b+1=0

Wy multiply elements

15b^2-10b^2-210b+1=0

We add all the numbers together, and all the variables

5b^2-210b+1=0

a = 5; b = -210; c = +1;
Δ = b2-4ac
Δ = -2102-4·5·1
Δ = 44080
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:

$\sqrt{\Delta}=\sqrt{44080}=\sqrt{16*2755}=\sqrt{16}*\sqrt{2755}=4\sqrt{2755}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-210)-4\sqrt{2755}}{2*5}=\frac{210-4\sqrt{2755}}{10}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-210)+4\sqrt{2755}}{2*5}=\frac{210+4\sqrt{2755}}{10}
$