3-3/7b=10.3b+10

Solution for 3-3/7b=10.3b+10 equation:

3-3/7b=10.3b+10

We move all terms to the left:

3-3/7b-(10.3b+10)=0


Domain of the equation: 7b!=0

b!=0/7

b!=0

b∈R


We get rid of parentheses

-3/7b-10.3b-10+3=0

We multiply all the terms by the denominator


-(10.3b)*7b-10*7b+3*7b-3=0

We add all the numbers together, and all the variables

-(+10.3b)*7b-10*7b+3*7b-3=0

We multiply parentheses

-70b^2-10*7b+3*7b-3=0

Wy multiply elements

-70b^2-70b+21b-3=0

We add all the numbers together, and all the variables

-70b^2-49b-3=0

a = -70; b = -49; c = -3;
Δ = b2-4ac
Δ = -492-4·(-70)·(-3)
Δ = 1561
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-49)-\sqrt{1561}}{2*-70}=\frac{49-\sqrt{1561}}{-140}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-49)+\sqrt{1561}}{2*-70}=\frac{49+\sqrt{1561}}{-140}
$