-4.9t^2+12.5t+105=0

Solution for -4.9t^2+12.5t+105=0 equation:

-4.9t^2+12.5t+105=0

a = -4.9; b = 12.5; c = +105;
Δ = b2-4ac
Δ = 12.52-4·(-4.9)·105
Δ = 2214.25
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{-(12.5)-\sqrt{2214.25}}{2*-4.9}=\frac{-12.5-\sqrt{2214.25}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12.5)+\sqrt{2214.25}}{2*-4.9}=\frac{-12.5+\sqrt{2214.25}}{-9.8}
$