-4.9t^2-4t+650=0

Solution for -4.9t^2-4t+650=0 equation:

-4.9t^2-4t+650=0

a = -4.9; b = -4; c = +650;
Δ = b2-4ac
Δ = -42-4·(-4.9)·650
Δ = 12756
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{12756}=\sqrt{4*3189}=\sqrt{4}*\sqrt{3189}=2\sqrt{3189}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-2\sqrt{3189}}{2*-4.9}=\frac{4-2\sqrt{3189}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2\sqrt{3189}}{2*-4.9}=\frac{4+2\sqrt{3189}}{-9.8}
$