1+3/5b=7/10b

Solution for 1+3/5b=7/10b equation:

1+3/5b=7/10b

We move all terms to the left:

1+3/5b-(7/10b)=0


Domain of the equation: 5b!=0

b!=0/5

b!=0

b∈R



Domain of the equation: 10b)!=0

b!=0/1

b!=0

b∈R


We add all the numbers together, and all the variables

3/5b-(+7/10b)+1=0

We get rid of parentheses

3/5b-7/10b+1=0

We calculate fractions


30b/50b^2+(-35b)/50b^2+1=0

We multiply all the terms by the denominator


30b+(-35b)+1*50b^2=0

Wy multiply elements

50b^2+30b+(-35b)=0

We get rid of parentheses

50b^2+30b-35b=0

We add all the numbers together, and all the variables

50b^2-5b=0

a = 50; b = -5; c = 0;
Δ = b2-4ac
Δ = -52-4·50·0
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-5}{2*50}=\frac{0}{100}
=0
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+5}{2*50}=\frac{10}{100}
=1/10
$