k^2-11k+23=0

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Solution for k^2-11k+23=0 equation: 



k^2-11k+23=0

a = 1; b = -11; c = +23;
Δ = b2-4ac
Δ = -112-4·1·23
Δ = 29
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-\sqrt{29}}{2*1}=\frac{11-\sqrt{29}}{2}
$


$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+\sqrt{29}}{2*1}=\frac{11+\sqrt{29}}{2}
$

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