3u^2-17u+16=0

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Solution for 3u^2-17u+16=0 equation:



3u^2-17u+16=0
a = 3; b = -17; c = +16;
Δ = b2-4ac
Δ = -172-4·3·16
Δ = 97
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-17)-\sqrt{97}}{2*3}=\frac{17-\sqrt{97}}{6} $
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-17)+\sqrt{97}}{2*3}=\frac{17+\sqrt{97}}{6} $

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