2x(4+x)+(3+x)=44

Solution for 2x(4+x)+(3+x)=44 equation:

2x(4+x)+(3+x)=44

We move all terms to the left:

2x(4+x)+(3+x)-(44)=0

We add all the numbers together, and all the variables

2x(x+4)+(x+3)-44=0

We multiply parentheses

2x^2+8x+(x+3)-44=0

We get rid of parentheses

2x^2+8x+x+3-44=0

We add all the numbers together, and all the variables

2x^2+9x-41=0

a = 2; b = 9; c = -41;
Δ = b2-4ac
Δ = 92-4·2·(-41)
Δ = 409
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{-(9)-\sqrt{409}}{2*2}=\frac{-9-\sqrt{409}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+\sqrt{409}}{2*2}=\frac{-9+\sqrt{409}}{4}
$