4.9t^2-17t-160=0

Solution for 4.9t^2-17t-160=0 equation:

4.9t^2-17t-160=0

a = 4.9; b = -17; c = -160;
Δ = b2-4ac
Δ = -172-4·4.9·(-160)
Δ = 3425
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:

$\sqrt{\Delta}=\sqrt{3425}=\sqrt{25*137}=\sqrt{25}*\sqrt{137}=5\sqrt{137}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-17)-5\sqrt{137}}{2*4.9}=\frac{17-5\sqrt{137}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-17)+5\sqrt{137}}{2*4.9}=\frac{17+5\sqrt{137}}{9.8}
$