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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-3F^2+4F-1=0

a = -3; b = 4; c = -1;
Δ = b2-4ac
Δ = 42-4·(-3)·(-1)
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2}{2*-3}=\frac{-6}{-6}
=1
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2}{2*-3}=\frac{-2}{-6}
=1/3
$