d^2+40d+8=0

Solution for d^2+40d+8=0 equation:

d^2+40d+8=0

a = 1; b = 40; c = +8;
Δ = b2-4ac
Δ = 402-4·1·8
Δ = 1568
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:

$\sqrt{\Delta}=\sqrt{1568}=\sqrt{784*2}=\sqrt{784}*\sqrt{2}=28\sqrt{2}$
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-28\sqrt{2}}{2*1}=\frac{-40-28\sqrt{2}}{2}
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+28\sqrt{2}}{2*1}=\frac{-40+28\sqrt{2}}{2}
$