1+3/10b=7/20b

Solution for 1+3/10b=7/20b equation:

1+3/10b=7/20b

We move all terms to the left:

1+3/10b-(7/20b)=0


Domain of the equation: 10b!=0

b!=0/10

b!=0

b∈R



Domain of the equation: 20b)!=0

b!=0/1

b!=0

b∈R


We add all the numbers together, and all the variables

3/10b-(+7/20b)+1=0

We get rid of parentheses

3/10b-7/20b+1=0

We calculate fractions


60b/200b^2+(-70b)/200b^2+1=0

We multiply all the terms by the denominator


60b+(-70b)+1*200b^2=0

Wy multiply elements

200b^2+60b+(-70b)=0

We get rid of parentheses

200b^2+60b-70b=0

We add all the numbers together, and all the variables

200b^2-10b=0

a = 200; b = -10; c = 0;
Δ = b2-4ac
Δ = -102-4·200·0
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-10}{2*200}=\frac{0}{400}
=0
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+10}{2*200}=\frac{20}{400}
=1/20
$