F(t)=-2t^2+22t

Solution for F(t)=-2t^2+22t equation:

(F)=-2F^2+22F

We move all terms to the left:

(F)-(-2F^2+22F)=0

We get rid of parentheses

2F^2-22F+F=0

We add all the numbers together, and all the variables

2F^2-21F=0

a = 2; b = -21; c = 0;
Δ = b2-4ac
Δ = -212-4·2·0
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{441}=21$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-21)-21}{2*2}=\frac{0}{4}
=0
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-21)+21}{2*2}=\frac{42}{4}
=10+1/2
$