3x^2-96x+92=0

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Solution for 3x^2-96x+92=0 equation:



3x^2-96x+92=0
a = 3; b = -96; c = +92;
Δ = b2-4ac
Δ = -962-4·3·92
Δ = 8112
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:
$\sqrt{\Delta}=\sqrt{8112}=\sqrt{2704*3}=\sqrt{2704}*\sqrt{3}=52\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-96)-52\sqrt{3}}{2*3}=\frac{96-52\sqrt{3}}{6} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-96)+52\sqrt{3}}{2*3}=\frac{96+52\sqrt{3}}{6} $

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