0=(2x+3)(x+1)

Solution for 0=(2x+3)(x+1) equation:

0=(2x+3)(x+1)

We move all terms to the left:

0-((2x+3)(x+1))=0

We add all the numbers together, and all the variables

-((2x+3)(x+1))=0

We multiply parentheses ..

-((+2x^2+2x+3x+3))=0


We calculate terms in parentheses: -((+2x^2+2x+3x+3)), so:

(+2x^2+2x+3x+3)

We get rid of parentheses

2x^2+2x+3x+3

We add all the numbers together, and all the variables

2x^2+5x+3

Back to the equation:

-(2x^2+5x+3)


We get rid of parentheses

-2x^2-5x-3=0

a = -2; b = -5; c = -3;
Δ = b2-4ac
Δ = -52-4·(-2)·(-3)
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{-(-5)-1}{2*-2}=\frac{4}{-4}
=-1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+1}{2*-2}=\frac{6}{-4}
=-1+1/2
$