4r^2-12r+8=0

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Solution for 4r^2-12r+8=0 equation: 



4r^2-12r+8=0

a = 4; b = -12; c = +8;
Δ = b2-4ac
Δ = -122-4·4·8
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{-(-12)-4}{2*4}=\frac{8}{8}
=1
$


$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4}{2*4}=\frac{16}{8}
=2
$

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