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16x^2+200x+1=0
a = 16; b = 200; c = +1;
Δ = b2-4ac
Δ = 2002-4·16·1
Δ = 39936
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{39936}=\sqrt{1024*39}=\sqrt{1024}*\sqrt{39}=32\sqrt{39}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(200)-32\sqrt{39}}{2*16}=\frac{-200-32\sqrt{39}}{32} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(200)+32\sqrt{39}}{2*16}=\frac{-200+32\sqrt{39}}{32} $
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