(5x+4)(x+1)=0

Solution for (5x+4)(x+1)=0 equation:

(5x+4)(x+1)=0

We multiply parentheses ..

(+5x^2+5x+4x+4)=0

We get rid of parentheses

5x^2+5x+4x+4=0

We add all the numbers together, and all the variables

5x^2+9x+4=0

a = 5; b = 9; c = +4;
Δ = b2-4ac
Δ = 92-4·5·4
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-1}{2*5}=\frac{-10}{10}
=-1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+1}{2*5}=\frac{-8}{10}
=-4/5
$