k^2-7k+6=0

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Solution for k^2-7k+6=0 equation: 



k^2-7k+6=0

a = 1; b = -7; c = +6;
Δ = b2-4ac
Δ = -72-4·1·6
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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


$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-5}{2*1}=\frac{2}{2}
=1
$


$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+5}{2*1}=\frac{12}{2}
=6
$

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