4.9t^2+340t-25=0

Solution for 4.9t^2+340t-25=0 equation:

4.9t^2+340t-25=0

a = 4.9; b = 340; c = -25;
Δ = b2-4ac
Δ = 3402-4·4.9·(-25)
Δ = 116090
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{-(340)-\sqrt{116090}}{2*4.9}=\frac{-340-\sqrt{116090}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(340)+\sqrt{116090}}{2*4.9}=\frac{-340+\sqrt{116090}}{9.8}
$