4.9t^2-14t-1.8=0

Solution for 4.9t^2-14t-1.8=0 equation:

4.9t^2-14t-1.8=0

a = 4.9; b = -14; c = -1.8;
Δ = b2-4ac
Δ = -142-4·4.9·(-1.8)
Δ = 231.28
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{-(-14)-\sqrt{231.28}}{2*4.9}=\frac{14-\sqrt{231.28}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+\sqrt{231.28}}{2*4.9}=\frac{14+\sqrt{231.28}}{9.8}
$