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k^2+6k-53=2
We move all terms to the left:
k^2+6k-53-(2)=0
We add all the numbers together, and all the variables
k^2+6k-55=0
a = 1; b = 6; c = -55;
Δ = b2-4ac
Δ = 62-4·1·(-55)
Δ = 256
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{256}=16$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-16}{2*1}=\frac{-22}{2} =-11 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+16}{2*1}=\frac{10}{2} =5 $
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