X^2+16x+41=10

Solution for X^2+16x+41=10 equation:

X^2+16X+41=10

We move all terms to the left:

X^2+16X+41-(10)=0

We add all the numbers together, and all the variables

X^2+16X+31=0

a = 1; b = 16; c = +31;
Δ = b2-4ac
Δ = 162-4·1·31
Δ = 132
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:

$\sqrt{\Delta}=\sqrt{132}=\sqrt{4*33}=\sqrt{4}*\sqrt{33}=2\sqrt{33}$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-2\sqrt{33}}{2*1}=\frac{-16-2\sqrt{33}}{2}
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+2\sqrt{33}}{2*1}=\frac{-16+2\sqrt{33}}{2}
$