Y=13t-4.9t^2

Solution for Y=13t-4.9t^2 equation:

=13Y-4.9Y^2

We move all terms to the left:

-(13Y-4.9Y^2)=0

We get rid of parentheses

4.9Y^2-13Y=0

a = 4.9; b = -13; c = 0;
Δ = b2-4ac
Δ = -132-4·4.9·0
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-13}{2*4.9}=\frac{0}{9.8}
=0
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+13}{2*4.9}=\frac{26}{9.8}
=2+1.8823529411765/2.8823529411765
$