3g^2-9g+6=0

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Solution for 3g^2-9g+6=0 equation: 



3g^2-9g+6=0

a = 3; b = -9; c = +6;
Δ = b2-4ac
Δ = -92-4·3·6
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$


$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-3}{2*3}=\frac{6}{6}
=1
$


$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+3}{2*3}=\frac{12}{6}
=2
$

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