80t+4.9t^2=96

Solution for 80t+4.9t^2=96 equation:

80t+4.9t^2=96

We move all terms to the left:

80t+4.9t^2-(96)=0

a = 4.9; b = 80; c = -96;
Δ = b2-4ac
Δ = 802-4·4.9·(-96)
Δ = 8281.6
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{-(80)-\sqrt{8281.6}}{2*4.9}=\frac{-80-\sqrt{8281.6}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(80)+\sqrt{8281.6}}{2*4.9}=\frac{-80+\sqrt{8281.6}}{9.8}
$