0=-16t^2+185

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Solution for 0=-16t^2+185 equation:



0=-16t^2+185
We move all terms to the left:
0-(-16t^2+185)=0
We add all the numbers together, and all the variables
-(-16t^2+185)=0
We get rid of parentheses
16t^2-185=0
a = 16; b = 0; c = -185;
Δ = b2-4ac
Δ = 02-4·16·(-185)
Δ = 11840
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:
$\sqrt{\Delta}=\sqrt{11840}=\sqrt{64*185}=\sqrt{64}*\sqrt{185}=8\sqrt{185}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{185}}{2*16}=\frac{0-8\sqrt{185}}{32} =-\frac{8\sqrt{185}}{32} =-\frac{\sqrt{185}}{4} $
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{185}}{2*16}=\frac{0+8\sqrt{185}}{32} =\frac{8\sqrt{185}}{32} =\frac{\sqrt{185}}{4} $

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