2r^2-8r+6=0

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Solution for 2r^2-8r+6=0 equation: 



2r^2-8r+6=0

a = 2; b = -8; c = +6;
Δ = b2-4ac
Δ = -82-4·2·6
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$


$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-4}{2*2}=\frac{4}{4}
=1
$


$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+4}{2*2}=\frac{12}{4}
=3
$

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