-4.9t^2+9t+400=0

Solution for -4.9t^2+9t+400=0 equation:

-4.9t^2+9t+400=0

a = -4.9; b = 9; c = +400;
Δ = b2-4ac
Δ = 92-4·(-4.9)·400
Δ = 7921
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{7921}=89$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-89}{2*-4.9}=\frac{-98}{-9.8}
=+10
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+89}{2*-4.9}=\frac{80}{-9.8}
=-8+1/6.125
$