-4.9t^2+2t+10=0

Solution for -4.9t^2+2t+10=0 equation:

-4.9t^2+2t+10=0

a = -4.9; b = 2; c = +10;
Δ = b2-4ac
Δ = 22-4·(-4.9)·10
Δ = 200
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:

$\sqrt{\Delta}=\sqrt{200}=\sqrt{100*2}=\sqrt{100}*\sqrt{2}=10\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-10\sqrt{2}}{2*-4.9}=\frac{-2-10\sqrt{2}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+10\sqrt{2}}{2*-4.9}=\frac{-2+10\sqrt{2}}{-9.8}
$