7=16t-4.9t^2

Solution for 7=16t-4.9t^2 equation:

7=16t-4.9t^2

We move all terms to the left:

7-(16t-4.9t^2)=0

We get rid of parentheses

4.9t^2-16t+7=0

a = 4.9; b = -16; c = +7;
Δ = b2-4ac
Δ = -162-4·4.9·7
Δ = 118.8
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{-(-16)-\sqrt{118.8}}{2*4.9}=\frac{16-\sqrt{118.8}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+\sqrt{118.8}}{2*4.9}=\frac{16+\sqrt{118.8}}{9.8}
$