0=(5x+4)(x+40)

Solution for 0=(5x+4)(x+40) equation:

0=(5x+4)(x+40)

We move all terms to the left:

0-((5x+4)(x+40))=0

We add all the numbers together, and all the variables

-((5x+4)(x+40))=0

We multiply parentheses ..

-((+5x^2+200x+4x+160))=0


We calculate terms in parentheses: -((+5x^2+200x+4x+160)), so:

(+5x^2+200x+4x+160)

We get rid of parentheses

5x^2+200x+4x+160

We add all the numbers together, and all the variables

5x^2+204x+160

Back to the equation:

-(5x^2+204x+160)


We get rid of parentheses

-5x^2-204x-160=0

a = -5; b = -204; c = -160;
Δ = b2-4ac
Δ = -2042-4·(-5)·(-160)
Δ = 38416
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{38416}=196$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-204)-196}{2*-5}=\frac{8}{-10}
=-4/5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-204)+196}{2*-5}=\frac{400}{-10}
=-40
$