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3x^2-32x+12=0
a = 3; b = -32; c = +12;
Δ = b2-4ac
Δ = -322-4·3·12
Δ = 880
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{880}=\sqrt{16*55}=\sqrt{16}*\sqrt{55}=4\sqrt{55}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-4\sqrt{55}}{2*3}=\frac{32-4\sqrt{55}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+4\sqrt{55}}{2*3}=\frac{32+4\sqrt{55}}{6} $
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