5t^2+9.8t-8=0

Solution for 5t^2+9.8t-8=0 equation:

5t^2+9.8t-8=0

a = 5; b = 9.8; c = -8;
Δ = b2-4ac
Δ = 9.82-4·5·(-8)
Δ = 256.04
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{-(9.8)-\sqrt{256.04}}{2*5}=\frac{-9.8-\sqrt{256.04}}{10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9.8)+\sqrt{256.04}}{2*5}=\frac{-9.8+\sqrt{256.04}}{10}
$