0=(2x+1)(x)

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

0=(2x+1)(x)

We move all terms to the left:

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

We add all the numbers together, and all the variables

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


We calculate terms in parentheses: -((2x+1)x), so:

(2x+1)x

We multiply parentheses

2x^2+x

Back to the equation:

-(2x^2+x)


We get rid of parentheses

-2x^2-x=0

We add all the numbers together, and all the variables

-2x^2-1x=0

a = -2; b = -1; c = 0;
Δ = b2-4ac
Δ = -12-4·(-2)·0
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{-(-1)-1}{2*-2}=\frac{0}{-4}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+1}{2*-2}=\frac{2}{-4}
=-1/2
$