0=d^2-12d+20

Solution for 0=d^2-12d+20 equation:

0=d^2-12d+20

We move all terms to the left:

0-(d^2-12d+20)=0

We add all the numbers together, and all the variables

-(d^2-12d+20)=0

We get rid of parentheses

-d^2+12d-20=0

We add all the numbers together, and all the variables

-1d^2+12d-20=0

a = -1; b = 12; c = -20;
Δ = b2-4ac
Δ = 122-4·(-1)·(-20)
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-8}{2*-1}=\frac{-20}{-2}
=+10
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+8}{2*-1}=\frac{-4}{-2}
=+2
$