(b+1)(4-b)=0

Solution for (b+1)(4-b)=0 equation:

(b+1)(4-b)=0

We add all the numbers together, and all the variables

(b+1)(-1b+4)=0

We multiply parentheses ..

(-1b^2+4b-1b+4)=0

We get rid of parentheses

-1b^2+4b-1b+4=0

We add all the numbers together, and all the variables

-1b^2+3b+4=0

a = -1; b = 3; c = +4;
Δ = b2-4ac
Δ = 32-4·(-1)·4
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-5}{2*-1}=\frac{-8}{-2}
=+4
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+5}{2*-1}=\frac{2}{-2}
=-1
$