-4.9t^2+16t+7=0

Solution for -4.9t^2+16t+7=0 equation:

-4.9t^2+16t+7=0

a = -4.9; b = 16; c = +7;
Δ = b2-4ac
Δ = 162-4·(-4.9)·7
Δ = 393.2
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{393.2}}{2*-4.9}=\frac{-16-\sqrt{393.2}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+\sqrt{393.2}}{2*-4.9}=\frac{-16+\sqrt{393.2}}{-9.8}
$