F(1)=-16t^2+45

Solution for F(1)=-16t^2+45 equation:

(1)=-16F^2+45

We move all terms to the left:

(1)-(-16F^2+45)=0

We get rid of parentheses

16F^2-45+1=0

We add all the numbers together, and all the variables

16F^2-44=0

a = 16; b = 0; c = -44;
Δ = b2-4ac
Δ = 02-4·16·(-44)
Δ = 2816
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:

$\sqrt{\Delta}=\sqrt{2816}=\sqrt{256*11}=\sqrt{256}*\sqrt{11}=16\sqrt{11}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{11}}{2*16}=\frac{0-16\sqrt{11}}{32}
=-\frac{16\sqrt{11}}{32}
=-\frac{\sqrt{11}}{2}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{11}}{2*16}=\frac{0+16\sqrt{11}}{32}
=\frac{16\sqrt{11}}{32}
=\frac{\sqrt{11}}{2}
$