k^2-13k+30=0

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Solution for k^2-13k+30=0 equation:



k^2-13k+30=0
a = 1; b = -13; c = +30;
Δ = b2-4ac
Δ = -132-4·1·30
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-7}{2*1}=\frac{6}{2} =3 $
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+7}{2*1}=\frac{20}{2} =10 $

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