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

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

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

a = -4.9; b = 10; c = +10;
Δ = b2-4ac
Δ = 102-4·(-4.9)·10
Δ = 296
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:

$\sqrt{\Delta}=\sqrt{296}=\sqrt{4*74}=\sqrt{4}*\sqrt{74}=2\sqrt{74}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{74}}{2*-4.9}=\frac{-10-2\sqrt{74}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{74}}{2*-4.9}=\frac{-10+2\sqrt{74}}{-9.8}
$