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3x^2-33x+12=0
a = 3; b = -33; c = +12;
Δ = b2-4ac
Δ = -332-4·3·12
Δ = 945
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{945}=\sqrt{9*105}=\sqrt{9}*\sqrt{105}=3\sqrt{105}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-33)-3\sqrt{105}}{2*3}=\frac{33-3\sqrt{105}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-33)+3\sqrt{105}}{2*3}=\frac{33+3\sqrt{105}}{6} $
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