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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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


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

(+2x^2+10x)

We get rid of parentheses

2x^2+10x

Back to the equation:

-(2x^2+10x)


We get rid of parentheses

-2x^2-10x=0

a = -2; b = -10; c = 0;
Δ = b2-4ac
Δ = -102-4·(-2)·0
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-10}{2*-2}=\frac{0}{-4}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+10}{2*-2}=\frac{20}{-4}
=-5
$