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}
$