9r^2-19r+9=0

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Solution for 9r^2-19r+9=0 equation: 



9r^2-19r+9=0

a = 9; b = -19; c = +9;
Δ = b2-4ac
Δ = -192-4·9·9
Δ = 37
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-\sqrt{37}}{2*9}=\frac{19-\sqrt{37}}{18}
$


$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+\sqrt{37}}{2*9}=\frac{19+\sqrt{37}}{18}
$

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