F(d)=36d^2+6d

Solution for F(d)=36d^2+6d equation:

(F)=36F^2+6F

We move all terms to the left:

(F)-(36F^2+6F)=0

We get rid of parentheses

-36F^2+F-6F=0

We add all the numbers together, and all the variables

-36F^2-5F=0

a = -36; b = -5; c = 0;
Δ = b2-4ac
Δ = -52-4·(-36)·0
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-5}{2*-36}=\frac{0}{-72}
=0
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+5}{2*-36}=\frac{10}{-72}
=-5/36
$