64t2+-84t+27=0

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Solution for 64t2+-84t+27=0 equation:



64t^2+-84t+27=0
We add all the numbers together, and all the variables
64t^2-84t=0
a = 64; b = -84; c = 0;
Δ = b2-4ac
Δ = -842-4·64·0
Δ = 7056
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{7056}=84$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-84)-84}{2*64}=\frac{0}{128} =0 $
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-84)+84}{2*64}=\frac{168}{128} =1+5/16 $

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