d^2-16d+29=0

Solution for d^2-16d+29=0 equation:

d^2-16d+29=0

a = 1; b = -16; c = +29;
Δ = b2-4ac
Δ = -162-4·1·29
Δ = 140
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{140}=\sqrt{4*35}=\sqrt{4}*\sqrt{35}=2\sqrt{35}$
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-2\sqrt{35}}{2*1}=\frac{16-2\sqrt{35}}{2}
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+2\sqrt{35}}{2*1}=\frac{16+2\sqrt{35}}{2}
$