F(t)=-2t^2+4t+3

Solution for F(t)=-2t^2+4t+3 equation:

(F)=-2F^2+4F+3

We move all terms to the left:

(F)-(-2F^2+4F+3)=0

We get rid of parentheses

2F^2-4F+F-3=0

We add all the numbers together, and all the variables

2F^2-3F-3=0

a = 2; b = -3; c = -3;
Δ = b2-4ac
Δ = -32-4·2·(-3)
Δ = 33
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-\sqrt{33}}{2*2}=\frac{3-\sqrt{33}}{4}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{33}}{2*2}=\frac{3+\sqrt{33}}{4}
$