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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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


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

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

We get rid of parentheses

2x^2+2x+5x+5

We add all the numbers together, and all the variables

2x^2+7x+5

Back to the equation:

-(2x^2+7x+5)


We get rid of parentheses

-2x^2-7x-5=0

a = -2; b = -7; c = -5;
Δ = b2-4ac
Δ = -72-4·(-2)·(-5)
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-3}{2*-2}=\frac{4}{-4}
=-1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+3}{2*-2}=\frac{10}{-4}
=-2+1/2
$