d^2+2d-5=0

Solution for d^2+2d-5=0 equation:

d^2+2d-5=0

a = 1; b = 2; c = -5;
Δ = b2-4ac
Δ = 22-4·1·(-5)
Δ = 24
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:

$\sqrt{\Delta}=\sqrt{24}=\sqrt{4*6}=\sqrt{4}*\sqrt{6}=2\sqrt{6}$
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{6}}{2*1}=\frac{-2-2\sqrt{6}}{2}
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{6}}{2*1}=\frac{-2+2\sqrt{6}}{2}
$