120+128t=16t^2

Solution for 120+128t=16t^2 equation:

120+128t=16t^2

We move all terms to the left:

120+128t-(16t^2)=0

determiningTheFunctionDomain
-16t^2+128t+120=0

a = -16; b = 128; c = +120;
Δ = b2-4ac
Δ = 1282-4·(-16)·120
Δ = 24064
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:

$\sqrt{\Delta}=\sqrt{24064}=\sqrt{256*94}=\sqrt{256}*\sqrt{94}=16\sqrt{94}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(128)-16\sqrt{94}}{2*-16}=\frac{-128-16\sqrt{94}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(128)+16\sqrt{94}}{2*-16}=\frac{-128+16\sqrt{94}}{-32}
$