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

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

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

We multiply parentheses ..

(+5b^2+15b-4b-12)=0

We get rid of parentheses

5b^2+15b-4b-12=0

We add all the numbers together, and all the variables

5b^2+11b-12=0

a = 5; b = 11; c = -12;
Δ = b2-4ac
Δ = 112-4·5·(-12)
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-19}{2*5}=\frac{-30}{10}
=-3
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+19}{2*5}=\frac{8}{10}
=4/5
$