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252-128x+12x^2=0
a = 12; b = -128; c = +252;
Δ = b2-4ac
Δ = -1282-4·12·252
Δ = 4288
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{4288}=\sqrt{64*67}=\sqrt{64}*\sqrt{67}=8\sqrt{67}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-128)-8\sqrt{67}}{2*12}=\frac{128-8\sqrt{67}}{24} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-128)+8\sqrt{67}}{2*12}=\frac{128+8\sqrt{67}}{24} $
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