d^2+28d+47=0

Solution for d^2+28d+47=0 equation:

d^2+28d+47=0

a = 1; b = 28; c = +47;
Δ = b2-4ac
Δ = 282-4·1·47
Δ = 596
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:

$\sqrt{\Delta}=\sqrt{596}=\sqrt{4*149}=\sqrt{4}*\sqrt{149}=2\sqrt{149}$
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-2\sqrt{149}}{2*1}=\frac{-28-2\sqrt{149}}{2}
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+2\sqrt{149}}{2*1}=\frac{-28+2\sqrt{149}}{2}
$