(t)=-16t^2+128t+50

Solution for (t)=-16t^2+128t+50 equation:

(t)=-16t^2+128t+50

We move all terms to the left:

(t)-(-16t^2+128t+50)=0

We get rid of parentheses

16t^2-128t+t-50=0

We add all the numbers together, and all the variables

16t^2-127t-50=0

a = 16; b = -127; c = -50;
Δ = b2-4ac
Δ = -1272-4·16·(-50)
Δ = 19329
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{-(-127)-\sqrt{19329}}{2*16}=\frac{127-\sqrt{19329}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-127)+\sqrt{19329}}{2*16}=\frac{127+\sqrt{19329}}{32}
$