4.9t^2+3t-14=0

Solution for 4.9t^2+3t-14=0 equation:

4.9t^2+3t-14=0

a = 4.9; b = 3; c = -14;
Δ = b2-4ac
Δ = 32-4·4.9·(-14)
Δ = 283.4
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{-(3)-\sqrt{283.4}}{2*4.9}=\frac{-3-\sqrt{283.4}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{283.4}}{2*4.9}=\frac{-3+\sqrt{283.4}}{9.8}
$