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

Solution for 44=(x+3)(2x+1) equation:

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

We move all terms to the left:

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

We multiply parentheses ..

-((+2x^2+x+6x+3))+44=0


We calculate terms in parentheses: -((+2x^2+x+6x+3)), so:

(+2x^2+x+6x+3)

We get rid of parentheses

2x^2+x+6x+3

We add all the numbers together, and all the variables

2x^2+7x+3

Back to the equation:

-(2x^2+7x+3)


We get rid of parentheses

-2x^2-7x-3+44=0

We add all the numbers together, and all the variables

-2x^2-7x+41=0

a = -2; b = -7; c = +41;
Δ = b2-4ac
Δ = -72-4·(-2)·41
Δ = 377
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-\sqrt{377}}{2*-2}=\frac{7-\sqrt{377}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+\sqrt{377}}{2*-2}=\frac{7+\sqrt{377}}{-4}
$