3x^2+6x-69=0

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Solution for 3x^2+6x-69=0 equation:



3x^2+6x-69=0
a = 3; b = 6; c = -69;
Δ = b2-4ac
Δ = 62-4·3·(-69)
Δ = 864
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:
$\sqrt{\Delta}=\sqrt{864}=\sqrt{144*6}=\sqrt{144}*\sqrt{6}=12\sqrt{6}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-12\sqrt{6}}{2*3}=\frac{-6-12\sqrt{6}}{6} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+12\sqrt{6}}{2*3}=\frac{-6+12\sqrt{6}}{6} $

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