-4.9t^2+3t+10=0

Solution for -4.9t^2+3t+10=0 equation:

-4.9t^2+3t+10=0

a = -4.9; b = 3; c = +10;
Δ = b2-4ac
Δ = 32-4·(-4.9)·10
Δ = 205
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{205}}{2*-4.9}=\frac{-3-\sqrt{205}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{205}}{2*-4.9}=\frac{-3+\sqrt{205}}{-9.8}
$