(b-5)(b-3)=0

Solution for (b-5)(b-3)=0 equation:

(b-5)(b-3)=0

We multiply parentheses ..

(+b^2-3b-5b+15)=0

We get rid of parentheses

b^2-3b-5b+15=0

We add all the numbers together, and all the variables

b^2-8b+15=0

a = 1; b = -8; c = +15;
Δ = b2-4ac
Δ = -82-4·1·15
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-2}{2*1}=\frac{6}{2}
=3
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+2}{2*1}=\frac{10}{2}
=5
$