F(t)=3t^2+6t

Solution for F(t)=3t^2+6t equation:

(F)=3F^2+6F

We move all terms to the left:

(F)-(3F^2+6F)=0

We get rid of parentheses

-3F^2+F-6F=0

We add all the numbers together, and all the variables

-3F^2-5F=0

a = -3; b = -5; c = 0;
Δ = b2-4ac
Δ = -52-4·(-3)·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*-3}=\frac{0}{-6}
=0
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+5}{2*-3}=\frac{10}{-6}
=-1+2/3
$