-27+3t^2=0

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Solution for -27+3t^2=0 equation: 



-27+3t^2=0

a = 3; b = 0; c = -27;
Δ = b2-4ac
Δ = 02-4·3·(-27)
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$


$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-18}{2*3}=\frac{-18}{6}
=-3
$


$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+18}{2*3}=\frac{18}{6}
=3
$

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