F(t)=-4t^2+16t+9

Solution for F(t)=-4t^2+16t+9 equation:

(F)=-4F^2+16F+9

We move all terms to the left:

(F)-(-4F^2+16F+9)=0

We get rid of parentheses

4F^2-16F+F-9=0

We add all the numbers together, and all the variables

4F^2-15F-9=0

a = 4; b = -15; c = -9;
Δ = b2-4ac
Δ = -152-4·4·(-9)
Δ = 369
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{369}=\sqrt{9*41}=\sqrt{9}*\sqrt{41}=3\sqrt{41}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-3\sqrt{41}}{2*4}=\frac{15-3\sqrt{41}}{8}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+3\sqrt{41}}{2*4}=\frac{15+3\sqrt{41}}{8}
$