4.9t^2+10t-100=0

Solution for 4.9t^2+10t-100=0 equation:

4.9t^2+10t-100=0

a = 4.9; b = 10; c = -100;
Δ = b2-4ac
Δ = 102-4·4.9·(-100)
Δ = 2060
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:

$\sqrt{\Delta}=\sqrt{2060}=\sqrt{4*515}=\sqrt{4}*\sqrt{515}=2\sqrt{515}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{515}}{2*4.9}=\frac{-10-2\sqrt{515}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{515}}{2*4.9}=\frac{-10+2\sqrt{515}}{9.8}
$