Y=4.9t^2+4t

Solution for Y=4.9t^2+4t equation:

=4.9Y^2+4Y

We move all terms to the left:

-(4.9Y^2+4Y)=0

We get rid of parentheses

-4.9Y^2-4Y=0

a = -4.9; b = -4; c = 0;
Δ = b2-4ac
Δ = -42-4·(-4.9)·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4}{2*-4.9}=\frac{0}{-9.8}
=0
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4}{2*-4.9}=\frac{8}{-9.8}
=-8/9.8
$