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16x^2-32x+4=0
a = 16; b = -32; c = +4;
Δ = b2-4ac
Δ = -322-4·16·4
Δ = 768
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{768}=\sqrt{256*3}=\sqrt{256}*\sqrt{3}=16\sqrt{3}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-16\sqrt{3}}{2*16}=\frac{32-16\sqrt{3}}{32} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+16\sqrt{3}}{2*16}=\frac{32+16\sqrt{3}}{32} $
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