y^2-16y+40=0

Solution for y^2-16y+40=0 equation:

y^2-16y+40=0

a = 1; b = -16; c = +40;
Δ = b2-4ac
Δ = -162-4·1·40
Δ = 96
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:

$\sqrt{\Delta}=\sqrt{96}=\sqrt{16*6}=\sqrt{16}*\sqrt{6}=4\sqrt{6}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-4\sqrt{6}}{2*1}=\frac{16-4\sqrt{6}}{2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+4\sqrt{6}}{2*1}=\frac{16+4\sqrt{6}}{2}
$