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

Solution for (35+b)(1+b)=0 equation:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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

We get rid of parentheses

b^2+b+35b+35=0

We add all the numbers together, and all the variables

b^2+36b+35=0

a = 1; b = 36; c = +35;
Δ = b2-4ac
Δ = 362-4·1·35
Δ = 1156
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1156}=34$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(36)-34}{2*1}=\frac{-70}{2}
=-35
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(36)+34}{2*1}=\frac{-2}{2}
=-1
$