F(t)=16t^2+45t+6

Solution for F(t)=16t^2+45t+6 equation:

(F)=16F^2+45F+6

We move all terms to the left:

(F)-(16F^2+45F+6)=0

We get rid of parentheses

-16F^2+F-45F-6=0

We add all the numbers together, and all the variables

-16F^2-44F-6=0

a = -16; b = -44; c = -6;
Δ = b2-4ac
Δ = -442-4·(-16)·(-6)
Δ = 1552
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1552}=\sqrt{16*97}=\sqrt{16}*\sqrt{97}=4\sqrt{97}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-44)-4\sqrt{97}}{2*-16}=\frac{44-4\sqrt{97}}{-32}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-44)+4\sqrt{97}}{2*-16}=\frac{44+4\sqrt{97}}{-32}
$