4.9*t^2-14t-17.5=0

Solution for 4.9*t^2-14t-17.5=0 equation:

4.9t^2-14t-17.5=0

a = 4.9; b = -14; c = -17.5;
Δ = b2-4ac
Δ = -142-4·4.9·(-17.5)
Δ = 539
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:

$\sqrt{\Delta}=\sqrt{539}=\sqrt{49*11}=\sqrt{49}*\sqrt{11}=7\sqrt{11}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-7\sqrt{11}}{2*4.9}=\frac{14-7\sqrt{11}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+7\sqrt{11}}{2*4.9}=\frac{14+7\sqrt{11}}{9.8}
$