(b+3)(b-3)=3b+1

Solution for (b+3)(b-3)=3b+1 equation:

(b+3)(b-3)=3b+1

We move all terms to the left:

(b+3)(b-3)-(3b+1)=0

We use the square of the difference formula

b^2-(3b+1)-9=0

We get rid of parentheses

b^2-3b-1-9=0

We add all the numbers together, and all the variables

b^2-3b-10=0

a = 1; b = -3; c = -10;
Δ = b2-4ac
Δ = -32-4·1·(-10)
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-7}{2*1}=\frac{-4}{2}
=-2
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+7}{2*1}=\frac{10}{2}
=5
$