16t2-9=0

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Solution for 16t2-9=0 equation:



16t^2-9=0
a = 16; b = 0; c = -9;
Δ = b2-4ac
Δ = 02-4·16·(-9)
Δ = 576
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{576}=24$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-24}{2*16}=\frac{-24}{32} =-3/4 $
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+24}{2*16}=\frac{24}{32} =3/4 $

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