7+16t-4.905t^2=0

Solution for 7+16t-4.905t^2=0 equation:

7+16t-4.905t^2=0

a = -4.905; b = 16; c = +7;
Δ = b2-4ac
Δ = 162-4·(-4.905)·7
Δ = 393.34
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{393.34}}{2*-4.905}=\frac{-16-\sqrt{393.34}}{-9.81}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+\sqrt{393.34}}{2*-4.905}=\frac{-16+\sqrt{393.34}}{-9.81}
$