-16t^2+64t+128=t=-3

Solution for -16t^2+64t+128=t=-3 equation:

-16t^2+64t+128=t=-3

We move all terms to the left:

-16t^2+64t+128-(t)=0

We add all the numbers together, and all the variables

-16t^2+63t+128=0

a = -16; b = 63; c = +128;
Δ = b2-4ac
Δ = 632-4·(-16)·128
Δ = 12161
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{-(63)-\sqrt{12161}}{2*-16}=\frac{-63-\sqrt{12161}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(63)+\sqrt{12161}}{2*-16}=\frac{-63+\sqrt{12161}}{-32}
$