F(t)=-5t^2+20+60

Solution for F(t)=-5t^2+20+60 equation:

(F)=-5F^2+20+60

We move all terms to the left:

(F)-(-5F^2+20+60)=0

We get rid of parentheses

5F^2+F-20-60=0

We add all the numbers together, and all the variables

5F^2+F-80=0

a = 5; b = 1; c = -80;
Δ = b2-4ac
Δ = 12-4·5·(-80)
Δ = 1601
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{-(1)-\sqrt{1601}}{2*5}=\frac{-1-\sqrt{1601}}{10}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{1601}}{2*5}=\frac{-1+\sqrt{1601}}{10}
$