5(b+3)(b-4)=0

Solution for 5(b+3)(b-4)=0 equation:

5(b+3)(b-4)=0

We multiply parentheses ..

5(+b^2-4b+3b-12)=0

We multiply parentheses

5b^2-20b+15b-60=0

We add all the numbers together, and all the variables

5b^2-5b-60=0

a = 5; b = -5; c = -60;
Δ = b2-4ac
Δ = -52-4·5·(-60)
Δ = 1225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1225}=35$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-35}{2*5}=\frac{-30}{10}
=-3
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+35}{2*5}=\frac{40}{10}
=4
$