x^2+16x-41=9

Solution for x^2+16x-41=9 equation:

x^2+16x-41=9

We move all terms to the left:

x^2+16x-41-(9)=0

We add all the numbers together, and all the variables

x^2+16x-50=0

a = 1; b = 16; c = -50;
Δ = b2-4ac
Δ = 162-4·1·(-50)
Δ = 456
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:

$\sqrt{\Delta}=\sqrt{456}=\sqrt{4*114}=\sqrt{4}*\sqrt{114}=2\sqrt{114}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-2\sqrt{114}}{2*1}=\frac{-16-2\sqrt{114}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+2\sqrt{114}}{2*1}=\frac{-16+2\sqrt{114}}{2}
$