3f^2-3=0

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Solution for 3f^2-3=0 equation: 



3f^2-3=0

a = 3; b = 0; c = -3;
Δ = b2-4ac
Δ = 02-4·3·(-3)
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$


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


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

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