3=-16t^2+50+0

Solution for 3=-16t^2+50+0 equation:

3=-16t^2+50+0

We move all terms to the left:

3-(-16t^2+50+0)=0

We get rid of parentheses

16t^2-50-0+3=0

We add all the numbers together, and all the variables

16t^2-47=0

a = 16; b = 0; c = -47;
Δ = b2-4ac
Δ = 02-4·16·(-47)
Δ = 3008
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:

$\sqrt{\Delta}=\sqrt{3008}=\sqrt{64*47}=\sqrt{64}*\sqrt{47}=8\sqrt{47}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{47}}{2*16}=\frac{0-8\sqrt{47}}{32}
=-\frac{8\sqrt{47}}{32}
=-\frac{\sqrt{47}}{4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{47}}{2*16}=\frac{0+8\sqrt{47}}{32}
=\frac{8\sqrt{47}}{32}
=\frac{\sqrt{47}}{4}
$