4d^2+16d+5=0

Solution for 4d^2+16d+5=0 equation:

4d^2+16d+5=0

a = 4; b = 16; c = +5;
Δ = b2-4ac
Δ = 162-4·4·5
Δ = 176
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:

$\sqrt{\Delta}=\sqrt{176}=\sqrt{16*11}=\sqrt{16}*\sqrt{11}=4\sqrt{11}$
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-4\sqrt{11}}{2*4}=\frac{-16-4\sqrt{11}}{8}
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+4\sqrt{11}}{2*4}=\frac{-16+4\sqrt{11}}{8}
$