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12x^2-104x+168=0
a = 12; b = -104; c = +168;
Δ = b2-4ac
Δ = -1042-4·12·168
Δ = 2752
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2752}=\sqrt{64*43}=\sqrt{64}*\sqrt{43}=8\sqrt{43}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-104)-8\sqrt{43}}{2*12}=\frac{104-8\sqrt{43}}{24} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-104)+8\sqrt{43}}{2*12}=\frac{104+8\sqrt{43}}{24} $
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