(b+3)(b+1)=35

Solution for (b+3)(b+1)=35 equation:

(b+3)(b+1)=35

We move all terms to the left:

(b+3)(b+1)-(35)=0

We multiply parentheses ..

(+b^2+b+3b+3)-35=0

We get rid of parentheses

b^2+b+3b+3-35=0

We add all the numbers together, and all the variables

b^2+4b-32=0

a = 1; b = 4; c = -32;
Δ = b2-4ac
Δ = 42-4·1·(-32)
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-12}{2*1}=\frac{-16}{2}
=-8
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+12}{2*1}=\frac{8}{2}
=4
$