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32x^2-18x+1=0
a = 32; b = -18; c = +1;
Δ = b2-4ac
Δ = -182-4·32·1
Δ = 196
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{196}=14$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-14}{2*32}=\frac{4}{64} =1/16 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+14}{2*32}=\frac{32}{64} =1/2 $
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