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k2+25=34
We move all terms to the left:
k2+25-(34)=0
We add all the numbers together, and all the variables
k^2-9=0
a = 1; b = 0; c = -9;
Δ = b2-4ac
Δ = 02-4·1·(-9)
Δ = 36
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{36}=6$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6}{2*1}=\frac{-6}{2} =-3 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6}{2*1}=\frac{6}{2} =3 $
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