F(t)=-2t^2+20t+11

Solution for F(t)=-2t^2+20t+11 equation:

(F)=-2F^2+20F+11

We move all terms to the left:

(F)-(-2F^2+20F+11)=0

We get rid of parentheses

2F^2-20F+F-11=0

We add all the numbers together, and all the variables

2F^2-19F-11=0

a = 2; b = -19; c = -11;
Δ = b2-4ac
Δ = -192-4·2·(-11)
Δ = 449
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{-(-19)-\sqrt{449}}{2*2}=\frac{19-\sqrt{449}}{4}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+\sqrt{449}}{2*2}=\frac{19+\sqrt{449}}{4}
$