T=13.7t+0.5-4.9t^2

Solution for T=13.7t+0.5-4.9t^2 equation:

=13.7T+0.5-4.9T^2

We move all terms to the left:

-(13.7T+0.5-4.9T^2)=0

We get rid of parentheses

4.9T^2-13.7T-0.5=0

a = 4.9; b = -13.7; c = -0.5;
Δ = b2-4ac
Δ = -13.72-4·4.9·(-0.5)
Δ = 197.49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$T_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$T_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$T_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13.7)-\sqrt{197.49}}{2*4.9}=\frac{13.7-\sqrt{197.49}}{9.8}
$
$T_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13.7)+\sqrt{197.49}}{2*4.9}=\frac{13.7+\sqrt{197.49}}{9.8}
$