3x^2+6x-0.5=0

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Solution for 3x^2+6x-0.5=0 equation: 



3x^2+6x-0.5=0

a = 3; b = 6; c = -0.5;
Δ = b2-4ac
Δ = 62-4·3·(-0.5)
Δ = 42
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-\sqrt{42}}{2*3}=\frac{-6-\sqrt{42}}{6}
$


$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+\sqrt{42}}{2*3}=\frac{-6+\sqrt{42}}{6}
$

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