3r2-11r+6=0

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Solution for 3r2-11r+6=0 equation: 



3r^2-11r+6=0

a = 3; b = -11; c = +6;
Δ = b2-4ac
Δ = -112-4·3·6
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$


$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-7}{2*3}=\frac{4}{6}
=2/3
$


$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+7}{2*3}=\frac{18}{6}
=3
$

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