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16x^2-192x-192=0
a = 16; b = -192; c = -192;
Δ = b2-4ac
Δ = -1922-4·16·(-192)
Δ = 49152
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{49152}=\sqrt{16384*3}=\sqrt{16384}*\sqrt{3}=128\sqrt{3}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-192)-128\sqrt{3}}{2*16}=\frac{192-128\sqrt{3}}{32} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-192)+128\sqrt{3}}{2*16}=\frac{192+128\sqrt{3}}{32} $
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