0=(2x+1)(x+5)

Solution for 0=(2x+1)(x+5) equation:

0=(2x+1)(x+5)

We move all terms to the left:

0-((2x+1)(x+5))=0

We add all the numbers together, and all the variables

-((2x+1)(x+5))=0

We multiply parentheses ..

-((+2x^2+10x+x+5))=0


We calculate terms in parentheses: -((+2x^2+10x+x+5)), so:

(+2x^2+10x+x+5)

We get rid of parentheses

2x^2+10x+x+5

We add all the numbers together, and all the variables

2x^2+11x+5

Back to the equation:

-(2x^2+11x+5)


We get rid of parentheses

-2x^2-11x-5=0

a = -2; b = -11; c = -5;
Δ = b2-4ac
Δ = -112-4·(-2)·(-5)
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-9}{2*-2}=\frac{2}{-4}
=-1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+9}{2*-2}=\frac{20}{-4}
=-5
$