3r2-11r+8=0

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Solution for 3r2-11r+8=0 equation: 



3r^2-11r+8=0

a = 3; b = -11; c = +8;
Δ = b2-4ac
Δ = -112-4·3·8
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$


$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-5}{2*3}=\frac{6}{6}
=1
$


$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+5}{2*3}=\frac{16}{6}
=2+2/3
$

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