30=-16t^2+128t

Solution for 30=-16t^2+128t equation:

30=-16t^2+128t

We move all terms to the left:

30-(-16t^2+128t)=0

We get rid of parentheses

16t^2-128t+30=0

a = 16; b = -128; c = +30;
Δ = b2-4ac
Δ = -1282-4·16·30
Δ = 14464
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:

$\sqrt{\Delta}=\sqrt{14464}=\sqrt{64*226}=\sqrt{64}*\sqrt{226}=8\sqrt{226}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-128)-8\sqrt{226}}{2*16}=\frac{128-8\sqrt{226}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-128)+8\sqrt{226}}{2*16}=\frac{128+8\sqrt{226}}{32}
$