(b+1)(b-6)=-12

Solution for (b+1)(b-6)=-12 equation:

(b+1)(b-6)=-12

We move all terms to the left:

(b+1)(b-6)-(-12)=0

We add all the numbers together, and all the variables

(b+1)(b-6)+12=0

We multiply parentheses ..

(+b^2-6b+b-6)+12=0

We get rid of parentheses

b^2-6b+b-6+12=0

We add all the numbers together, and all the variables

b^2-5b+6=0

a = 1; b = -5; c = +6;
Δ = b2-4ac
Δ = -52-4·1·6
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-1}{2*1}=\frac{4}{2}
=2
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+1}{2*1}=\frac{6}{2}
=3
$