0=-16t2+6

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Solution for 0=-16t2+6 equation:



0=-16t^2+6
We move all terms to the left:
0-(-16t^2+6)=0
We add all the numbers together, and all the variables
-(-16t^2+6)=0
We get rid of parentheses
16t^2-6=0
a = 16; b = 0; c = -6;
Δ = b2-4ac
Δ = 02-4·16·(-6)
Δ = 384
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:
$\sqrt{\Delta}=\sqrt{384}=\sqrt{64*6}=\sqrt{64}*\sqrt{6}=8\sqrt{6}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{6}}{2*16}=\frac{0-8\sqrt{6}}{32} =-\frac{8\sqrt{6}}{32} =-\frac{\sqrt{6}}{4} $
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{6}}{2*16}=\frac{0+8\sqrt{6}}{32} =\frac{8\sqrt{6}}{32} =\frac{\sqrt{6}}{4} $

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