6r2-8r+1=0

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Solution for 6r2-8r+1=0 equation:



6r^2-8r+1=0
a = 6; b = -8; c = +1;
Δ = b2-4ac
Δ = -82-4·6·1
Δ = 40
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:
$\sqrt{\Delta}=\sqrt{40}=\sqrt{4*10}=\sqrt{4}*\sqrt{10}=2\sqrt{10}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-2\sqrt{10}}{2*6}=\frac{8-2\sqrt{10}}{12} $
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+2\sqrt{10}}{2*6}=\frac{8+2\sqrt{10}}{12} $

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