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12x^2-128x+240=0
a = 12; b = -128; c = +240;
Δ = b2-4ac
Δ = -1282-4·12·240
Δ = 4864
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{4864}=\sqrt{256*19}=\sqrt{256}*\sqrt{19}=16\sqrt{19}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-128)-16\sqrt{19}}{2*12}=\frac{128-16\sqrt{19}}{24} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-128)+16\sqrt{19}}{2*12}=\frac{128+16\sqrt{19}}{24} $
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