2+3/5(b)=7/10(b)

Solution for 2+3/5(b)=7/10(b) equation:

2+3/5(b)=7/10(b)

We move all terms to the left:

2+3/5(b)-(7/10(b))=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)+2=0

We get rid of parentheses

3/5b-7/10b+2=0

We calculate fractions


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

We multiply all the terms by the denominator


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

Wy multiply elements

100b^2+30b+(-35b)=0

We get rid of parentheses

100b^2+30b-35b=0

We add all the numbers together, and all the variables

100b^2-5b=0

a = 100; b = -5; c = 0;
Δ = b2-4ac
Δ = -52-4·100·0
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*100}=\frac{0}{200}
=0
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+5}{2*100}=\frac{10}{200}
=1/20
$