F(1)=-4.9t^2+3825

Solution for F(1)=-4.9t^2+3825 equation:

(1)=-4.9F^2+3825

We move all terms to the left:

(1)-(-4.9F^2+3825)=0

We get rid of parentheses

4.9F^2-3825+1=0

We add all the numbers together, and all the variables

4.9F^2-3824=0

a = 4.9; b = 0; c = -3824;
Δ = b2-4ac
Δ = 02-4·4.9·(-3824)
Δ = 74950.4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-\sqrt{74950.4}}{2*4.9}=\frac{0-\sqrt{74950.4}}{9.8}
=-\frac{\sqrt{}}{9.8}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+\sqrt{74950.4}}{2*4.9}=\frac{0+\sqrt{74950.4}}{9.8}
=\frac{\sqrt{}}{9.8}
$