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k2-6k-65=-6
We move all terms to the left:
k2-6k-65-(-6)=0
We add all the numbers together, and all the variables
k^2-6k-59=0
a = 1; b = -6; c = -59;
Δ = b2-4ac
Δ = -62-4·1·(-59)
Δ = 272
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}$$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{272}=\sqrt{16*17}=\sqrt{16}*\sqrt{17}=4\sqrt{17}$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-4\sqrt{17}}{2*1}=\frac{6-4\sqrt{17}}{2} $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+4\sqrt{17}}{2*1}=\frac{6+4\sqrt{17}}{2} $
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