2r^2-6r+4=0

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Solution for 2r^2-6r+4=0 equation: 



2r^2-6r+4=0

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

$\sqrt{\Delta}=\sqrt{4}=2$


$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2}{2*2}=\frac{4}{4}
=1
$


$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2}{2*2}=\frac{8}{4}
=2
$

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