2b+7b2-2+3b=1+b

Solution for 2b+7b2-2+3b=1+b equation:

2b+7b^2-2+3b=1+b

We move all terms to the left:

2b+7b^2-2+3b-(1+b)=0

We add all the numbers together, and all the variables

7b^2+2b+3b-(b+1)-2=0

We add all the numbers together, and all the variables

7b^2+5b-(b+1)-2=0

We get rid of parentheses

7b^2+5b-b-1-2=0

We add all the numbers together, and all the variables

7b^2+4b-3=0

a = 7; b = 4; c = -3;
Δ = b2-4ac
Δ = 42-4·7·(-3)
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{-(4)-10}{2*7}=\frac{-14}{14}
=-1
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+10}{2*7}=\frac{6}{14}
=3/7
$