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16x^2-128x+200=0
a = 16; b = -128; c = +200;
Δ = b2-4ac
Δ = -1282-4·16·200
Δ = 3584
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{3584}=\sqrt{256*14}=\sqrt{256}*\sqrt{14}=16\sqrt{14}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-128)-16\sqrt{14}}{2*16}=\frac{128-16\sqrt{14}}{32} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-128)+16\sqrt{14}}{2*16}=\frac{128+16\sqrt{14}}{32} $
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