2x^2+11x-41=x

Solution for 2x^2+11x-41=x equation:

2x^2+11x-41=x

We move all terms to the left:

2x^2+11x-41-(x)=0

We add all the numbers together, and all the variables

2x^2+10x-41=0

a = 2; b = 10; c = -41;
Δ = b2-4ac
Δ = 102-4·2·(-41)
Δ = 428
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{428}=\sqrt{4*107}=\sqrt{4}*\sqrt{107}=2\sqrt{107}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{107}}{2*2}=\frac{-10-2\sqrt{107}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{107}}{2*2}=\frac{-10+2\sqrt{107}}{4}
$