-4.9t^2+6t+10=0

Solution for -4.9t^2+6t+10=0 equation:

-4.9t^2+6t+10=0

a = -4.9; b = 6; c = +10;
Δ = b2-4ac
Δ = 62-4·(-4.9)·10
Δ = 232
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{232}=\sqrt{4*58}=\sqrt{4}*\sqrt{58}=2\sqrt{58}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{58}}{2*-4.9}=\frac{-6-2\sqrt{58}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{58}}{2*-4.9}=\frac{-6+2\sqrt{58}}{-9.8}
$