-4.9t^2+3.62t+0.75=0

Solution for -4.9t^2+3.62t+0.75=0 equation:

-4.9t^2+3.62t+0.75=0

a = -4.9; b = 3.62; c = +0.75;
Δ = b2-4ac
Δ = 3.622-4·(-4.9)·0.75
Δ = 27.8044
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{-(3.62)-\sqrt{27.8044}}{2*-4.9}=\frac{-3.62-\sqrt{27.8044}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3.62)+\sqrt{27.8044}}{2*-4.9}=\frac{-3.62+\sqrt{27.8044}}{-9.8}
$