0=d^2+5d-4

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

0=d^2+5d-4

We move all terms to the left:

0-(d^2+5d-4)=0

We add all the numbers together, and all the variables

-(d^2+5d-4)=0

We get rid of parentheses

-d^2-5d+4=0

We add all the numbers together, and all the variables

-1d^2-5d+4=0

a = -1; b = -5; c = +4;
Δ = b2-4ac
Δ = -52-4·(-1)·4
Δ = 41
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-\sqrt{41}}{2*-1}=\frac{5-\sqrt{41}}{-2}
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{41}}{2*-1}=\frac{5+\sqrt{41}}{-2}
$