16t-0.3t^2=120

Solution for 16t-0.3t^2=120 equation:

16t-0.3t^2=120

We move all terms to the left:

16t-0.3t^2-(120)=0

a = -0.3; b = 16; c = -120;
Δ = b2-4ac
Δ = 162-4·(-0.3)·(-120)
Δ = 112
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:

$\sqrt{\Delta}=\sqrt{112}=\sqrt{16*7}=\sqrt{16}*\sqrt{7}=4\sqrt{7}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-4\sqrt{7}}{2*-0.3}=\frac{-16-4\sqrt{7}}{-0.6}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+4\sqrt{7}}{2*-0.3}=\frac{-16+4\sqrt{7}}{-0.6}
$