Y=-4.9t^2+3t+9

Solution for Y=-4.9t^2+3t+9 equation:

=-4.9Y^2+3Y+9

We move all terms to the left:

-(-4.9Y^2+3Y+9)=0

We get rid of parentheses

4.9Y^2-3Y-9=0

a = 4.9; b = -3; c = -9;
Δ = b2-4ac
Δ = -32-4·4.9·(-9)
Δ = 185.4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-\sqrt{185.4}}{2*4.9}=\frac{3-\sqrt{185.4}}{9.8}
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{185.4}}{2*4.9}=\frac{3+\sqrt{185.4}}{9.8}
$