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12x^2-38x+12=0
a = 12; b = -38; c = +12;
Δ = b2-4ac
Δ = -382-4·12·12
Δ = 868
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{868}=\sqrt{4*217}=\sqrt{4}*\sqrt{217}=2\sqrt{217}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-38)-2\sqrt{217}}{2*12}=\frac{38-2\sqrt{217}}{24} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-38)+2\sqrt{217}}{2*12}=\frac{38+2\sqrt{217}}{24} $
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