4.9t^2-19t-70=0

Solution for 4.9t^2-19t-70=0 equation:

4.9t^2-19t-70=0

a = 4.9; b = -19; c = -70;
Δ = b2-4ac
Δ = -192-4·4.9·(-70)
Δ = 1733
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-\sqrt{1733}}{2*4.9}=\frac{19-\sqrt{1733}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+\sqrt{1733}}{2*4.9}=\frac{19+\sqrt{1733}}{9.8}
$