4.9t^2+5.1t-105=0

Solution for 4.9t^2+5.1t-105=0 equation:

4.9t^2+5.1t-105=0

a = 4.9; b = 5.1; c = -105;
Δ = b2-4ac
Δ = 5.12-4·4.9·(-105)
Δ = 2084.01
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{-(5.1)-\sqrt{2084.01}}{2*4.9}=\frac{-5.1-\sqrt{2084.01}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5.1)+\sqrt{2084.01}}{2*4.9}=\frac{-5.1+\sqrt{2084.01}}{9.8}
$