# 0=-16t^2+221

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## Solution for 0=-16t^2+221 equation:

0=-16t^2+221
We move all terms to the left:
0-(-16t^2+221)=0
We add all the numbers together, and all the variables
-(-16t^2+221)=0
We get rid of parentheses
16t^2-221=0
a = 16; b = 0; c = -221;Δ = b2-4acΔ = 02-4·16·(-221)Δ = 14144The delta value is higher than zero, so the equation has two solutionsWe use following formulas to calculate our solutions:$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$The end solution:
$\sqrt{\Delta}=\sqrt{14144}=\sqrt{64*221}=\sqrt{64}*\sqrt{221}=8\sqrt{221}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{221}}{2*16}=\frac{0-8\sqrt{221}}{32} =-\frac{8\sqrt{221}}{32} =-\frac{\sqrt{221}}{4}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{221}}{2*16}=\frac{0+8\sqrt{221}}{32} =\frac{8\sqrt{221}}{32} =\frac{\sqrt{221}}{4}$

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