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k2+26k+17=0
We add all the numbers together, and all the variables
k^2+26k+17=0
a = 1; b = 26; c = +17;
Δ = b2-4ac
Δ = 262-4·1·17
Δ = 608
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}$$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{608}=\sqrt{16*38}=\sqrt{16}*\sqrt{38}=4\sqrt{38}$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(26)-4\sqrt{38}}{2*1}=\frac{-26-4\sqrt{38}}{2} $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(26)+4\sqrt{38}}{2*1}=\frac{-26+4\sqrt{38}}{2} $
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