2u^2-6u+1=0

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Solution for 2u^2-6u+1=0 equation:



2u^2-6u+1=0
a = 2; b = -6; c = +1;
Δ = b2-4ac
Δ = -62-4·2·1
Δ = 28
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:
$\sqrt{\Delta}=\sqrt{28}=\sqrt{4*7}=\sqrt{4}*\sqrt{7}=2\sqrt{7}$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{7}}{2*2}=\frac{6-2\sqrt{7}}{4} $
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{7}}{2*2}=\frac{6+2\sqrt{7}}{4} $

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