-16t^2+128t=-30

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

-16t^2+128t=-30

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

a = -16; b = 128; c = +30;
Δ = b2-4ac
Δ = 1282-4·(-16)·30
Δ = 18304
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:

$\sqrt{\Delta}=\sqrt{18304}=\sqrt{64*286}=\sqrt{64}*\sqrt{286}=8\sqrt{286}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(128)-8\sqrt{286}}{2*-16}=\frac{-128-8\sqrt{286}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(128)+8\sqrt{286}}{2*-16}=\frac{-128+8\sqrt{286}}{-32}
$