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

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

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

We add all the numbers together, and all the variables

(-1b+3)(5b+4)=0

We multiply parentheses ..

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

We get rid of parentheses

-5b^2-4b+15b+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
$