2x(x+1)+x=41

Solution for 2x(x+1)+x=41 equation:

2x(x+1)+x=41

We move all terms to the left:

2x(x+1)+x-(41)=0

We add all the numbers together, and all the variables

x+2x(x+1)-41=0

We multiply parentheses

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

We add all the numbers together, and all the variables

2x^2+3x-41=0

a = 2; b = 3; c = -41;
Δ = b2-4ac
Δ = 32-4·2·(-41)
Δ = 337
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{-(3)-\sqrt{337}}{2*2}=\frac{-3-\sqrt{337}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{337}}{2*2}=\frac{-3+\sqrt{337}}{4}
$