y^2-16y-41=0

Solution for y^2-16y-41=0 equation:

y^2-16y-41=0

a = 1; b = -16; c = -41;
Δ = b2-4ac
Δ = -162-4·1·(-41)
Δ = 420
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{420}=\sqrt{4*105}=\sqrt{4}*\sqrt{105}=2\sqrt{105}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-2\sqrt{105}}{2*1}=\frac{16-2\sqrt{105}}{2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+2\sqrt{105}}{2*1}=\frac{16+2\sqrt{105}}{2}
$