-4.9t^2+24t+8=0

Solution for -4.9t^2+24t+8=0 equation:

-4.9t^2+24t+8=0

a = -4.9; b = 24; c = +8;
Δ = b2-4ac
Δ = 242-4·(-4.9)·8
Δ = 732.8
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{-(24)-\sqrt{732.8}}{2*-4.9}=\frac{-24-\sqrt{732.8}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+\sqrt{732.8}}{2*-4.9}=\frac{-24+\sqrt{732.8}}{-9.8}
$