(b-5)(b+12)=0

Solution for (b-5)(b+12)=0 equation:

(b-5)(b+12)=0

We multiply parentheses ..

(+b^2+12b-5b-60)=0

We get rid of parentheses

b^2+12b-5b-60=0

We add all the numbers together, and all the variables

b^2+7b-60=0

a = 1; b = 7; c = -60;
Δ = b2-4ac
Δ = 72-4·1·(-60)
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-17}{2*1}=\frac{-24}{2}
=-12
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+17}{2*1}=\frac{10}{2}
=5
$