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8x^2-62x+16=0
a = 8; b = -62; c = +16;
Δ = b2-4ac
Δ = -622-4·8·16
Δ = 3332
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{3332}=\sqrt{196*17}=\sqrt{196}*\sqrt{17}=14\sqrt{17}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-62)-14\sqrt{17}}{2*8}=\frac{62-14\sqrt{17}}{16} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-62)+14\sqrt{17}}{2*8}=\frac{62+14\sqrt{17}}{16} $
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