T=-4.9t^2+15t

Solution for T=-4.9t^2+15t equation:

=-4.9T^2+15T

We move all terms to the left:

-(-4.9T^2+15T)=0

We get rid of parentheses

4.9T^2-15T=0

a = 4.9; b = -15; c = 0;
Δ = b2-4ac
Δ = -152-4·4.9·0
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$T_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$T_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$T_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-15}{2*4.9}=\frac{0}{9.8}
=0
$
$T_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+15}{2*4.9}=\frac{30}{9.8}
=3+0.6/9.8
$