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

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

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

We multiply parentheses ..

(-5x^2-15x-4x-12)=0

We get rid of parentheses

-5x^2-15x-4x-12=0

We add all the numbers together, and all the variables

-5x^2-19x-12=0

a = -5; b = -19; c = -12;
Δ = b2-4ac
Δ = -192-4·(-5)·(-12)
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-11}{2*-5}=\frac{8}{-10}
=-4/5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+11}{2*-5}=\frac{30}{-10}
=-3
$